MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 12

A mass of 0.2kg is attached to the lower end of a massless spring of force-constant 200 N/m, the upper end of which is fixed to a rigid support. Which of the following statements is/are true?

0%
In equilibrium, the spring will be stretched by 1cm.
0%
If the mass is raised till the spring is unstretched state and then released, it will go down by 2cm

before moving upwards.
0%
The frequency of oscillation will be nearly 5 Hz
0%
If the system is taken to the moon, the frequency of oscillation will be the same as on the earth.

A particle of mass 200 g executes linear simple harmonic motion with an amplitude 10 cm. When the particles at a point midway between the mean and the extreme position, its kinetic energy is

$3\pi ^{2}\times 10^{-3}J$ . Assuming the initial phase to be

$\dfrac{2\pi }{3}$ , the equation of motion of the particle will be :

0%
y=10 sin $(2\pi t +\dfrac{2\pi }{3})$ cm
0%
y=10 sin $(4\pi t +\dfrac{2\pi }{3})$ cm
0%
y=10 cos $(2\pi t +\dfrac{\pi }{6})$ cm
0%
y=10 cos $(2\pi t +\dfrac{\pi }{3})$ cm

A block rides on a piston that is moving vertically with simple harmonic motion. The maximum speed of the piston is

$2m/s$ . At what amplitude of motion will the block and piston separate?

$(g=10m/s^{2})$

0%
$20 cm$
0%
$30 cm$
0%
$40 cm$
0%
$50 cm$

Explanation

Let

$A$ be amplitude
Now, maximum speed of piston

$= 2 m/s$

$\therefore w A = 2 m/s$
If

$w^{2} A = g$ then they will separate.

$\Rightarrow 2\times w = g$

$\Rightarrow w = 5 rad/sec$

$\therefore A = \dfrac{2}{5} m = 40 cm$

The angular frequency of a spring block system is

$\omega _{0}$ This system is suspended from the ceiling of anelevator moving downwards with a constant speed v

$_{0}$ . The block is at rest relative to the elevator. Lift issuddenly stopped. Assuming the downwards as a positive direction, choose the wrong statement:

0%
The amplitude of the block is $\dfrac{v_{0}}{\omega_{0}}$
0%
The initial phase of the block is $\pi$
0%
The equation of motion for the block is $\dfrac{v_{0}}{\omega_{0}} \sin \omega _{0}$ t.
0%
The maximum speed of the block is v $_{0}$ .

Explanation

Since

${ V }_{ 0 }$ is the maximum speed of the ball in its equilibrium position.

$\therefore { V }_{ 0 }={ \omega }_{ 0 }A\\ A=\dfrac { { V }_{ 0 } }{ { \omega }_{ 0 } }$

So equation of motion becomes

$\dfrac { { V }_{ 0 } }{ { \omega }_{ 0 } } =Asin{ \omega }_{ 0 }t$ as there was no initial phase in the SHM.

There was no initial phase in the SHM.

Hence B is correct.

A mass

$m$ , which is attached to a spring with constant

$k$ , oscillates on a horizontal table, with amplitude

$A$ . At an instant when the spring is stretched by

${\sqrt{3}A}/{2}$ , a second mass

$m$ is dropped vertically onto the original mass and immediately sticks to it. What is the amplitude of the resulting motion?

0%
$\displaystyle\frac{\sqrt{3}}{2}A$
0%
$\displaystyle\sqrt{\frac{7}{8}}A$
0%
$\displaystyle\sqrt{\frac{13}{16}}A$
0%
$\displaystyle\sqrt{\frac{2}{3}}A$

Explanation

Formula for the velocity

$v$ , of a particle/mass executing SHM with amplitude

$A$ , and at displacement

$x$ is,

$v=\omega \sqrt{A^2-x^2}$
now at

$x= \dfrac{A \sqrt{3}}{2}$ , velocity will be,

$\dfrac{A \omega}{2}$ , at this point another mass

$m$ attached to the mass vertically preserving linear momentum. that causes for the whole mass move with a new velocity

$\dfrac{\omega A}{4}$ at

$\dfrac{A \sqrt{3}}{2}$ and with a different angular velocity

$\omega/\sqrt{2}$ as we know

$\omega=\sqrt{\dfrac{k}{m}}$ , mass doubles in the second situation.
Substituting new values of velocity and angular frequency in the equation of velocity at displacement

$\dfrac{A \sqrt{3}}{2}$ , new amplitude turns out to be

$A \sqrt{\dfrac{7}{8}}$

A particle of mass

$m$ is moving in a field where the potential energy is given by

$U(x) = U_0(1 - \cos \:ax)$ , where

$U_0$ and

$a$ are positive constants and

$x$ is the displacement from mean position. Then (for small oscillations):

0%
the time period is $T=2\pi \sqrt{\dfrac{m}{aU_0}}$
0%
the speed of the particle is maximum at $x = 0$
0%
the amplitude of oscillations is $\dfrac{\pi }{a}$
0%
the time period is $T=2\pi \sqrt{\dfrac{m}{a^2U_0}}$

Explanation

$U(x) = U_{0}(1-cos ax)$

$\dfrac{dU}{dx}= U_{0}a sin ax$

$F =- dU/dx$

$F = - U_{0}a sin ax$
for small value of "x" we will write,

$F = - U_{0} a^{2} x$

$a=- U_{0} a^{2}x /m$

$T = 2 \pi \sqrt{ \dfrac{m}{ U_{0}a^{2}}}$
at x = 0, speed of particle will be maximum

A particle of mass 10 gm is placed in a potential field given by

$V = (50x^2 + 100) \ J/kg$ . The frequency of oscillation in cycles/sec is :

0%
$\dfrac{10}{\pi}$
0%
$\dfrac{5}{\pi}$
0%
$\dfrac{100}{\pi}$
0%
$\dfrac{50}{\pi}$

Explanation

Potential energy

$U = mV$

$\Rightarrow U = (50x^2+100)10^{-2}$

$F = -\dfrac{dU}{dx} = -(100x) 10^{-2}$

$\Rightarrow m{\omega}^2x = -(100\times 10^{-2})x$

$10\times 10^{-3}{\omega}^2x = 100\times 10^{-2}x$

$\Rightarrow {\omega}^2 = 100, {\omega} = 10$

$\Rightarrow f = \dfrac{\omega}{2\pi}=\dfrac{10}{2\pi}=\dfrac{5}{\pi}$

A particle which can move along x-axis is released from rest at the position

$x = x_0.$ The potential energy (U) of the block is described below :

$U=\begin{Bmatrix} -ax &;x<0 \\ bx^2 &;x\geq0 \end{Bmatrix}$
Which of the following statements is/are correct :

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The subsequent motion is simple harmonic
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The subsequent motion is periodic
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The speed is a continuous function of time
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The magnitude of acceleration is a continuous function of time

Explanation

The potential is not quadratic so the motion is not sinusoidal or simple harmonic. The motion is periodic since the particle is bounded by the potential. The energy is constant and so the particle will move back and forth between a maximum negative position and the position

$x=x_0.$
The potential energy is continuous and hence

$K.E.=TE-U$ is continuous. Therefore, speed is continuous.
The derivative of the potential energy is not continuous at x = 0. Since, force is equal to the negative derivative of the potential energy, the force is not continuous at

$x=0$ . Therefore, acceleration

$(\frac{F}{m})$ is not continuous at

$x=0$ .

A canon of mass

$1 kg$ (including the bullet) is connected with a spring of a spring constant

$k = 100 N/m$ , and is performing SHM with an amplitude of

$1 m$ . When the cannon reaches

$x = \dfrac{\sqrt{3}} {2}m$ from the equilibrium, moving along +x direction, a bullet of

$\dfrac{1} {2}kg$ is suddenly fired from it with a velocity of

$20 m/s$ relative to the ground. What will be the new amplitude of the cannon ?

0%
$\dfrac{\sqrt{3}} {2}$
0%
$\dfrac{\sqrt{5}} {2}$
0%
$\dfrac{\sqrt{7}} {2}$
0%
$\dfrac{\sqrt{11}} {2}$

Explanation

Initial angular frequency

$\omega_{n} = \sqrt{\dfrac{k} {m}} = \sqrt{\dfrac{100} {1}} = 10 sec^{-1}$
and velocity

$v(x=\dfrac{\sqrt{3}}{2}) = \omega_{n} \sqrt{A^{2} - x^{2}} = 10\sqrt{1^{2} - \left( \sqrt{3}/2 \right )^{2}}=5m/s$
Velocity of the cannon after firing the bullet can be found using momentum conservation

$\left ( 1 \right ) \left ( 5 \right ) = \left ( \dfrac{1} {2} \right ) \left ( 20 \right ) + \left ( \dfrac{1} {2} \right ) v \\ \Rightarrow v = -10m/s$
lets say new amplitude is

$A_1$ and new angular frequency is

$\omega_{n}$ , then we have

$v = \omega_{n} \sqrt{A_{1}^{2} - x^{2}} \Rightarrow -10$

$= \sqrt{\dfrac{100} {\left ( 1/2 \right )}} \sqrt{A_{1}^{2} - \left ( \dfrac{\sqrt{3}} {2} \right )^{2}} \Rightarrow A_{1} = \dfrac{\sqrt{5}} {2} m$

A wedge of mass

$m$ , having a smooth semi circular part of radius R is resting on smooth horizontal surface.Now a particle of mass

$m/2$ is released from the top point of the semicircular part. Then,

0%
The maximum displacement of the wedge will be R
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The maximum displacement of the wedge will be $\dfrac{2R} {3}$
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The wedge will perform simple harmonic motion of amplitude R
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The wedge will perform oscillatory motion of amplitude $\dfrac{R} {3}$

Explanation

As there is no net external force acting on the system in horizontal direction , momentum is conserved in this direction and equal to zero .
Hence , displacement of center of mass is zero .

Displacement of center of mass is given by

$\left ( S_{x} \right )_{cm} = \dfrac{m_{1} {S_{x}} _{1} + m_{2} {S_{x} }_{2}}{m_{1} + m_{2}}$

$0 = \dfrac{m_{1} x + m_{2} (x-2R)}{m_{1} + m_{2}}$
Hence ,

$x=\dfrac{2R}{3}$
As the mean position is taken as zero displacement and the maximum displacement is

$\dfrac{2R}{3}$ , the wedge is in an oscillatory motion of amplitude

$\dfrac{R}{3}$ about its mean position.

A block of mass

$m$ is attached to the spring

$k$ in free length and released at time

$t=0$ in the position

$O$ from rest. For the subsequent vertical motion of the block the equation of motion is given by

$\left ( \omega =\sqrt{\displaystyle \frac{k}{m}} \right )$

0%
$x=\displaystyle \frac{mg}{k}\sin \omega t$
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$x=\displaystyle \frac{mg}{k}\cos \omega t$
0%
$x=\displaystyle \frac{mg}{k}(1+ \cos \omega t)$
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$x=\displaystyle \frac{mg}{k}(1- \cos \omega t)$

Explanation

The angular velocity of the system is

$\omega=\sqrt{k/m}$

Here, the point

$O$ is the extreme point of SHM.

The amplitude,

$A$ is given by,

$kA=mg\Rightarrow A=mg/k$

So the equation of motion from point

$O$ is

$A-Asin(\omega t+\pi/2)=\dfrac{mg}{k}(1-cos\omega t)$

A particle is an linear simple harmonic motion between two extreme point A and B. 10 cm apart(See figure below) If the direction from A to B is taken as positive direction, what are signs of displacement x, velocity V and acceleration a, when the particle is at A?

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$x=-ve, V=-ve, a=-ve$
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$x=+ve, V=0, a=-ve$
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$x=+ve, V=-ve, a=+ve$
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$x=-ve, V=0, a=+ve$

A particle in SHM has an amplitude of 20 cm and time period of 2 sec. Its maximum velocity will be(in m/s):

0%
$0.04 \ \pi^2$
0%
$0.2 \ \pi$
0%
$0.2 \ \pi^2$
0%
$0.04 \ \pi$

Explanation

$v=\dfrac{dx}{dt}=A\omega cos(\omega t + \theta)$

Maximum $v$ occurs for maximum value of $cos(\omega t + \theta)$ , which is 1
Thus, maximum $v=A\omega$
Now, $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{2}=\pi$ , and $A = 20 cm = 0.2 m$
Thus, maximum v = $0.2\pi$

The potential energy of a particle of mass

$0.1\ kg$ moving along x-axis, is given by

$U=5x(x-4)\ J$ , where

$x$ is in meters. It can be concluded that

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the particle is acted upon by a constant force
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the speed of the particle is maximum at $x=2\ m$
0%
the particle executes simple harmonic motion
0%
the period of oscillation of the particle is ${\pi/5 s}$

Explanation

$F=-\dfrac { dU }{ dx } =-\dfrac { d }{ dx } \left( 5{ x }^{ 2 }-20x \right) =-10x+20$
Here we see that force is directly proportional to the displacement and opposite to the direction of displacement. So it satisfies the condition of SHM.

$\Rightarrow$ A is incorrect and C is correct.
Speed of the particle is maximum in mean position. i.e. where force is zero. i.e.

$F=-10x+20=0 \Rightarrow x=2m$
Speed of the particle is maximum at

$x=2m$ ..

$\Rightarrow$ B is correct.
Since

$F=-10x+20$ comparing it with general equation of SHM

$F=-kx$ , we have

$k=10N/m$

$\Rightarrow T=2\pi \sqrt { \dfrac { m }{ k } }$

$\Rightarrow T=2\pi \sqrt { \dfrac { 0.1 }{ 10 } }=2\pi \sqrt { \dfrac { 1 }{ 100 } }=\dfrac{2\pi}{10}=\dfrac{\pi}{5}s$

$\Rightarrow$ D is correct.

The figure shows a graph between velocity and displacement (from mean position) of a particle performing SHM:

0%
The time period of the particle is 1.57s
0%
The maximum acceleration will be ${40 cm /s^{2}}$
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The velocity of a particle is ${2\sqrt21 cm/s}$ when it is at a distance of 1 cm from the mean position
0%
None of these

Explanation

by graph we can see that

$V_{max}=10cm/s$ and

$A=2.5cm$ and we know that

$V_{max}=A \omega$

$\Rightarrow 10=2.5 \omega$

$\Rightarrow \omega=4s^{-1}$

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{4}=\dfrac{3.14}{2}=1.57s$ $\Rightarrow$ A correct

$a_{max}=A \omega^2=2.5 \times 4^2=40cms^{-2}$ $\Rightarrow$ B correct

$v(x=1cm)=w \sqrt{A^2-x^2}=4\sqrt{2.5^2-1^2}=4\sqrt{5.25}=4\sqrt{\dfrac{525}{100}}=4\sqrt{\dfrac{21\times 25}{100}}=2\sqrt{21} cm/s$ $\Rightarrow$ C correct & D wrong.

Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations (i), (ii) and (iii) are in the ratio

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$1:\sqrt{2}:1/\sqrt{2}$
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$2:\sqrt{2}:1/\sqrt{2}$
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$1/\sqrt{2}:2:1$
0%
$2:1/\sqrt{2}:1$

Explanation

By above formulae, we get,

$T_{1}=2\pi \sqrt{\frac{m}{k}}$

$T_{2}=2\pi \sqrt{\frac{2m}{k}}$ ..(Series combination of springs)

$T_{3}=2\pi \sqrt{\frac{m}{2k}}$ ..(Parallel combination of springs)
Taking ratio, we get Answer as option A.

PE of a particle is

$\displaystyle U(x)=\frac{a}{x^2}-\frac{b}{x}$ . Find the time period of small oscillation

0%
$\displaystyle 2\pi\sqrt\frac{8a^3m}{b^4}$
0%
$\displaystyle 2\pi\sqrt\frac{8b^3m}{b^4}$
0%
$\displaystyle 2\pi\sqrt\frac{8a^4m}{b^3}$
0%
$\displaystyle 2\pi\sqrt\frac{8b^4m}{a^3}$

Explanation

The equilibrium position is

$\displaystyle \frac{du}{dx}=0$

$\displaystyle \frac{-2a}{x_0^3}+\frac{b}{x_0^2}=0$
or

$\displaystyle x_0=\frac{2a}{b}$

$\displaystyle u(x)=\frac{a}{x_0^2}-\frac{b}{x_0}+\left| (x-x_0)\frac{dU}{dx}\right |_{x=x_0}+\left |\frac{1}{2}(x-x_0)^2\frac{d^2u}{dx^2}\right|_{x=x_0}$

$\displaystyle \left|\frac{d^2u}{dx^2}\right|_{x=x_0}=\frac{6a}{x_0^4}-\frac{2b}{x_0^3}=\frac{6a}{\left(\displaystyle \frac{2a}{b}\right)^4}-\frac{2b}{\left(\displaystyle \frac{2a}{b}\right)^3}=\frac{b^4}{8a^3}$
Thus

$\displaystyle u(x)=U(x_0)+\frac{1}{2}\left (\frac{b^4}{8a^3} \right) y^2$
comparing with

$\displaystyle \frac{1}{2}m\omega^2 y^2=\frac{1}{2}\frac{b^4}{8a^3} y^2$

$\displaystyle \omega = \sqrt{\frac{b^4}{8a^3m}}$
or

$\displaystyle T=2\pi\sqrt{\frac{8a^3m}{b^4}}$

A hydrogen atom has mass

$1.68\times 10^{-27}$ kg. When attached to a certain massive molecule it oscillates with a frequency

$10^{14}$ Hz and with an amplitude

$10^{-9}$ cm. Find the force acting on the hydrogen atom.

0%
$2.21\times 10^{-9}$ N
0%
$3.31\times 10^{-9}$ N
0%
$4.42\times 10^{-9}$ N
0%
$6.63\times 10^{-9}$ N

Explanation

$\omega ^{ 2 }=\dfrac { k }{ m } \\ \Rightarrow { \left( 2\pi f \right) }^{ 2 }=\dfrac { k }{ m } \\ \Rightarrow k=4\pi ^{ 2 }f^{ 2 }m$
and

$F=kx_{ 0 }=4\pi ^{ 2 }f^{ 2 }mx_{ 0 }\\ \Rightarrow F=4\times \pi ^{ 2 }\times 10^{ 28 }\times 1.68\times 10^{ -27 }\times 10^{ -11 }N\\ \Rightarrow F=6.63\times 10^{ -9 }N$

A particle of mass

$m$ moves according to the equation

$F=-amr$ where

$a$ is a positive constant

$, r$ is radius vector.

$r=r_0\hat{i}$ and

$v=v_0\hat{j}$ at

$t=0$ . Describe the trajectory.

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$\displaystyle\left(\dfrac{x}{r_0}\right)^2+a\left(\dfrac{y}{v_0}\right)^2=1$
0%
$\displaystyle\left(\dfrac{x}{r_0}\right)^2+a\left(\dfrac{y}{v_0}\right)^2=0$
0%
$\displaystyle\left(\dfrac{x}{r_0}\right)^2+\left(\dfrac{y}{v_0}\right)^2=\frac{1}{\alpha}$
0%
$none \quad of \quad these$

Explanation

$F=-amr\\ \Rightarrow F_{ x }\hat { i } +F_{ y }\hat { j } =-am\left( x\hat { i } +y\hat { j } \right) \\ \Rightarrow m\dfrac { d^{ 2 }x }{ dt^{ 2 } } \hat { i } +m\dfrac { d^{ 2 }y }{ dt^{ 2 } } \hat { j } =-am\left( x\hat { i } +y\hat { j } \right)$

$\dfrac { d^{ 2 }x }{ dt^{ 2 } } =-ax$ and

$\dfrac { d^{ 2 }y }{ dt^{ 2 } } =-ay\quad$ ...............................................................

$(i)$

Thus

$x=x_o\sin{(\omega t+\theta)}$ and

$\omega^2=a$ .........................................................................................

$(ii)$

$\dfrac{dx}{dt}=x_o\omega\cos{(\omega t+\theta)}$
since

$v_x=0$ when

$t=0$

$\Rightarrow x_o\omega\cos{(\theta)}=0$

$\Rightarrow \theta=\dfrac{\pi}{2}$

$\Rightarrow x=x_o \sin\left( \omega t+\dfrac { \pi }{ 2 } \right)$

$x=x_o \cos\left( \omega t \right)$

$\Rightarrow x=r_o \cos\left( \omega t \right)$ ...............................................

$(iii)$

$\because x(t=0)=r_o$

Now from

$(i)$ , we have,

$y=y_0\sin{(\omega t+\beta)}$

$y=y_0\sin{(\omega t)} \because y(t=0)=0$

$\dfrac{dy}{dt}=y_0\omega\cos {(\omega t)}$
since

$v_y=v_o$ when

$t=0$

$\Rightarrow v_o=y_0\omega$

$\Rightarrow y=\dfrac{v_o}{\omega}\sin{(\omega t)}$ .....

$(iv)$

from

$(iii)$ and

$(iv)$ we have,

$\left(\dfrac{x}{r_0}\right)^2+\left(\dfrac{y}{v_0/\omega}\right)^2=1$

$\Rightarrow \left(\dfrac{x}{r_0}\right)^2+\omega^2\left(\frac{y}{v_0}\right)^2=1$

$\Rightarrow \left(\dfrac{x}{r_0}\right)^2+a\left(\dfrac{y}{v_0}\right)^2=1 \because \omega^2=a$ from

$(ii)$

Two masses

$m_1$ and

$m_2$ are connected to a spring of spring constant

$K$ at two ends. The spring is compressed by

$y$ and released. The distance moved by

$m_1$ before it comes to a stop for the first time is

0%
$\displaystyle \dfrac{m_1y}{m_1+m_2}$
0%
$\displaystyle \dfrac{m_2y}{m_1+m_2}$
0%
$\displaystyle \dfrac{2m_1y}{m_1+m_2}$
0%
$\displaystyle \dfrac{2m_2y}{m_1+m_2}$

Explanation

the spring is compressed by y therefore the force on masses is

$F = Ky$

let

$m_1$ traveled distance a and mass

$m_2$ be b

therefore

$F = ma$

$F(t) = - m_1 \dfrac{da(t)^2}{dt^2}$

similarly for second mass

$-F(t) = m_2\dfrac{db(t)^2}{dt^2}$

Now Assuming simple harmonic motion

$a(t) =Asin \omega t$ and

$b(t) = B sin \omega t$

as the spring force is equal for both masses

$m_1a = m_2b$

$m_1 A sin \omega t = m_2 B sin \omega t$

$m_!A = m_2 B$

As long as the above amplitude equation will hold center of mass wont move

Eq for center of mass is

$0 =\dfrac{-Bm_2 + Am_1}{m_1+m_2}$

we know

$A+B = y$

$m_1A = m_2(y-A)$

$A = \dfrac{m_2y}{m_1+m_2}$

In the figure shown, a spring mass system is placed on a horizontal smooth surface in between two vertical rigid walls

$W_{1}$ and

$W_{2}$ . One end of spring is fixed with wall

$W_{1}$ and other end is attached with mass

$m$ which is free to move. Initially, spring is tension free and having natural length

$l_{0}$ . Mass

$m$ is compressed through a distance a and released. Taking the collision between wall

$W_{2}$ and mass

$m$ as elastic and

$K$ as spring constant, the average force exerted by mass

$m$ on wall

$W_{2}$ in one oscillation of block is

0%
$\displaystyle \frac{2aK}{\pi }$
0%
$\displaystyle \frac{2ma}{\pi }$
0%
$\displaystyle \frac{aK}{\pi }$
0%
$\displaystyle \frac{2aK}{m}$

Explanation

maximum velocity

$\displaystyle v_{max}=\omega A\left ( \sqrt{\frac{K}{m}} \right )a$
Change in momentum

$\Delta P=2mv_{max}=2m\left ( \sqrt{\frac{K}{m}} \right )a$
Force exerted by the mass

$m$ on wall

$W_2$ is

$\displaystyle F=\frac{\Delta P}{\Delta t}=\frac{\Delta P}{T/2}=\frac{2ma\sqrt{k/m}}{\pi \sqrt{m/k}}=\frac{2aK}{\pi }$

The time period of the oscillating system (image above) is

0%
$T = 2\pi\sqrt{\displaystyle\frac{m}{k_1k_2}}$
0%
$T = 2\pi\sqrt{\displaystyle\frac{m}{k_1+k_2}}$
0%
$T = 2\pi\sqrt{mk_1k_2}$
0%
None of these

Explanation

Let block is displaced by

$X$ to the right.

Restoring Force by spring 1,

$F_1=k_1X$

Restoring Force by spring 2,

$F_2=k_2X$

Net Restoring force

$=(k_1+k_2)X$

$\Rightarrow m\ddot{X}= - (k_1+k_2)X$

$\Rightarrow \ddot{X}= -\dfrac{(k_1+ k_2)}{m} X$

$\Rightarrow \ddot{X}= -\omega^2 X$

$\Rightarrow \omega= \sqrt{\dfrac{(k_1+k_2)}{m}}$

$\therefore \text{Time Period } T= \dfrac{2\pi}{\omega}$

$\Rightarrow T= 2\pi \sqrt{\dfrac{m}{k_1+k_2}}$

A particle of mass is executing oscillations about the origin on the x-axis. Its potential energy is

$V(x) = k|x|^3$ , where

$k$ is a positive constant. If the amplitude of oscillation is

$a$ , then its time period

$T$ is proportional

0%
proportional to $\displaystyle\frac{1}{\sqrt{a}}$
0%
proportional to $\sqrt{a}$
0%
Independent $a^{\frac{3}{2}}$
0%
None of these

Explanation

$\quad V(x) = k|x|^3$
Since,

$\quad F = -\displaystyle\frac{dV(x)}{dx} = -3k|x|^2 \quad ...(i)$

$\quad x = a\sin(\omega t)$
This equation always fits to the differential equation

$\quad \displaystyle\frac{d^2x}{dt^2} = -\omega^2x$ or

$m\displaystyle\frac{d^2x}{dt} = -m\omega^2x$

$\quad \Rightarrow F = - m\omega^2x\quad ...(ii)$
Equations

$(i)$ and

$(ii)$ give

$\quad -3k|x|^2 = -m\omega^2x$

$\quad \Rightarrow \omega = \sqrt{\displaystyle\frac{3kx}{m}} = \sqrt{\displaystyle\frac{3ka}{m}}[\sin(\omega t)]^{1/2}$

$\quad\Rightarrow \omega \propto \sqrt{a}$

$\quad \Rightarrow T \propto \displaystyle\frac{1}{\sqrt{a}}$

Two spring-mass systems support equal mass and have spring constants

$\displaystyle K_{1}$ and

$\displaystyle K_{2}$ . If the maximum velocities in two systems are equal then ratio of amplitude of 1st to that of 2nd is

0%
$\displaystyle \sqrt{K_{1}/K_{1}}$
0%
$\displaystyle K_{1}/K_{2}$
0%
$\displaystyle K_{2}/K_{1}$
0%
$\displaystyle \sqrt{K_{2}/K_{1}}$

Explanation

Maximum velocity

$V_{max}=\omega A$

$\omega=\sqrt{\frac{K}{m}}$

$V_{1}=\sqrt{\dfrac{K_{1}}{m}}A_{1}$

$V_{2}=\sqrt{\dfrac{K_{2}}{m}}A_{2}$

It is given that both have same maximum velocity and same mass

$V_{1}=V_{2}$

$\sqrt{\dfrac{K_{1}}{m}}A_{1}=\sqrt{\dfrac{K_{2}}{m}}A_{2}$

$\dfrac{A_{1}}{A_{2}}=\sqrt{\dfrac{K_{2}}{K_{1}}}$

A spring has natural length

$40$ cm and spring constant

$500 N/m$ . A block of mass

$1 kg$ is attached at one end of the spring and other end of the spring is attached to a ceiling. The block is released from the position, where the spring has length

$45 cm$ :

0%
the block will perform SHM of amplitude $5 cm$
0%
the block will have maximum velocity $30\sqrt{5}$ cm/s
0%
the block will have maximum acceleration 15 m/s $^{2}$
0%
the minimum elastic potential energy of the spring will be zero

Explanation

$kx_{0}=mg$

$\therefore$

$\displaystyle x_{0}=\frac{mg}{k}=\frac{1\times 10}{500}$

$=0.02$ m

$=2$ cm
So equilibrium is obtained after an extension of 2 cm of at a length of 42 cm. But it is released from a length of 45 cm.

$\therefore$ A=3 cm=0.03 m
(b)

$\displaystyle v_{max}=\omega A=\sqrt{\frac{k}{m}}A$

$\displaystyle =\left ( \sqrt{\frac{500}{1}} \right )\left ( 0.03 \right )=0.3\sqrt{5}$ m/s

$=30\sqrt{5}$ cm/s
(c)

$a_{max}=\omega ^{2}A$

$\displaystyle =\left ( \frac{k}{m} \right )\left ( A \right )=\left ( \frac{500}{1} \right )\left ( 0.03 \right )=15$ m/s

$^{2}$
(d) Mean position is at 42 cm length and amplitude is 3 cm. Hence block oscillates between 45 cm length and 39 cm. Natural length 40 cm lies in between these two, where elastic potential energy=0.

One end of an ideal spring is fixed to a wall at origin

$O$ and axis of spring is parallel to

$x-$ axis. A block of mass

$m=1 \ kg$ is attached to free end of the spring and it is performing

$SHM$ . Equation of position of the block in coordinate system shown in figure is

$x=10+3\sin 10t$ . Here,

$t$ is in second and

$x$ in

$cm$ . Another block of mass

$M=3\ kg$ , moving towards the origin with velocity

$30\ cm/s$ collides with the block performing

$SHM$ at

$t=0$ and gets stuck to it calculate.
(i) New amplitude of oscillation
(ii) New equation for position of the combined body.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

From equation

$x=10+3sin(10t)$

$\omega=10$ and

$A=3$

We know

$\omega=\dfrac{k}{m}$

$k=10^2\times 1=100N/m$

The velocity of mass m at t=0 is

$v_1=\omega\times A=10\times 3=30cm/s$

Now from momentum conservation

$(M+m)v=30\times M-v_1\times m$

$4v=30\times 3-30\times 1$

$v=15cm/s$

Now to find the new amplitude of system

Let the new amplitude be

$A_1$

By conservation of energy

$\dfrac{(M+m)v^2}{2}=\dfrac{kA_1^2}{2}$

$A_1=0.03m=3cm$

and new angular velocity of system

$\omega_1=\sqrt{\dfrac{k}{M+m}}$

$\omega=\sqrt{\dfrac{100}{4}}=5rad/s$

So the equation is

$x'=10-3sin5t$

$x=3$

An elastic string of length

$\displaystyle l$ supports a heavy particle of mass m and the system is in equilibrium with elongation produced being e as shown in the figure. The particle is now pulled down below the equilibrium position through a distance

$\displaystyle d\left ( \leq e \right )$ and released. The angular frequency and maximum amplitude for SHM is

0%
$\displaystyle \sqrt{\frac{g}{e}},e$
0%
$\displaystyle \sqrt{\frac{g}{l }},2e$
0%
$\displaystyle \sqrt{\frac{g}{d+e}},d$
0%
$\displaystyle \sqrt{\frac{g}{e}},2d$

Explanation

Here, weight of the ball must balance spring force at equilibrium position.

$kx=mg\\ ke=mg\\ k=\frac { mg }{ e }$

After elongating more, the net restoring force will be,

$F=kd\\ a=\frac { Kd }{ m } \\ a={ \omega }^{ 2 }x\\ \omega =\sqrt { \frac { k }{ m } } \\ =\sqrt { \frac { mg }{ em } } \\ =\sqrt { \frac { g }{ e } } \\ T=\frac { 2\pi }{ \omega } =2\pi \sqrt { \frac { m }{ k } }$

Now, putting the value of K

$T=2\pi \sqrt { \frac { m\times e }{ mg } } \\ T=2\pi \sqrt { \frac { e }{ g } } \\ g={ \omega }^{ 2 }A$ [At extreme position]

$g=\frac { g }{ e } \times A\\ A=e$

A practice is executing SHM. its time period is equal to the smallest time interval in which the particle acquires a particular velocity,

$\bar { v }$ , the magnitude of

$\bar { v }$ may be:

0%
Zero
0%
$\displaystyle V_{max}$
0%
$\displaystyle \frac{V_{max}}{2}$
0%
$\displaystyle \frac{V_{max}}{\sqrt{2}}$

A particle oscillates simple harmonically with a period of 16 s. Two second after crossing the equilibrium position its velocity becomes 1 m/s. The amplitude is

0%
$\displaystyle \frac{\pi }{4}$ m
0%
$\displaystyle \frac{8\sqrt{2}}{\pi }$ m
0%
$\displaystyle \frac{8}{\pi }$ m
0%
$\displaystyle \frac{4\sqrt{2}}{\pi }$ m

Explanation

$\displaystyle \omega =\frac{2\pi }{T}=\frac{2\pi }{16}=\frac{\pi }{8}$ rad/s
At t=0, particle crosses the mean position. Hence its velocity is maximum. So, velocity as a function of time can be written as

$v=v_{max}\cos \omega t$
or

$v=\omega A\cos \omega t$

$\therefore$

$\displaystyle 1=\left ( \frac{\pi }{8} \right )A\cos \left ( \frac{\pi }{8} \right )\left ( 2 \right )=\left ( \frac{\pi }{8\sqrt{2}} \right )A$

$\therefore$

$\displaystyle A=\frac{8\sqrt{2}}{\pi }$ m

A particle moves along y-axis according to the equation y (in cm)

$=3\sin 100\pi t+8\sin ^{2}50\pi t-6$

0%
the particle performs SHM
0%
the amplitude of the particle's oscillation is 5 cm
0%
the mean position of the particle is at y=- 2 cm
0%
the particle does not perform SHM

Explanation

The given equation can be written as

$y=3\sin 100\pi t+\left ( 4-4\cos 100\pi t \right )-6$

$=3\sin 100\pi t+4\sin \left ( 100\pi t+\pi \pi /2 \right )-2$
or

$y=5\sin \left ( 100\pi -53^{\circ} \right )-2$

$y_{max}=5-2=3$

$y_{min}=-5-2=-7$
Mean position

$\displaystyle =\frac{y_{max}+y_{min}}{2}=-2$ cm

A particle is performing SHM of amplitude "A" and time period "T". Find the time taken by the particle to go from 0 to A /

$\sqrt{2}$ .

0%
T/12
0%
T/ $\sqrt{2}$
0%
T/6
0%
t/4

Explanation

Let equation of SHM be x = A

$\displaystyle \sin \omega t$
when x = 0 t = 0 when x = A/2 ; A/2 = A

$\displaystyle \sin \omega t$
or

$\displaystyle \sin \omega t=1/2$

$\displaystyle \omega t=\pi /6$

$\displaystyle \frac{2\pi }{T}t=\pi /6$ t=T/12
Hence time taken is T/12 where T is time period of SHM

Tow identical springs are connected to mass

$m$ as shown (

$k=$ spring constant). IF the period of the configuration in (a) is

$2s$ , the period of the configuration in (b) is

0%
$\sqrt {2}s$
0%
$1s$
0%
$\cfrac{1}{\sqrt {2}}s$
0%
$2\sqrt {2}s$

Explanation

Series combination :

Equivalent spring constant:

$\Rightarrow k_s = \dfrac{k_1 \times k_2}{k_1+k_2} = \dfrac{k \times k}{k+k} = \dfrac{k}{2}$

Time period,

$T_s =2\pi \sqrt{\dfrac{m}{k_s}} = 2\pi \sqrt{\dfrac{2m}{k}}$ ............(1)

Parallel combination :

Equivalent spring constant:

$\Rightarrow k_p = k_1 +k_2 = k+k = 2k$

Time period,

$T_p =2\pi \sqrt{\dfrac{m}{k_p}} = 2\pi \sqrt{\dfrac{m}{2k}}$ ............(2)

$\Rightarrow$

$\dfrac{T_p}{T_s} = \dfrac{1/\sqrt{2}}{\sqrt{2}} = \dfrac{1}{2}$

$\Rightarrow \dfrac{T_p}{2} = \dfrac{1}{2}$

$\Rightarrow T_p = 1\ s$

A particle executing simple harmonic motion has an angular frequency of 6.28

$\displaystyle s^{-1}$ and amplitude of 10 cm, find the speed at t = 1/6 s assuming that the motion starts from rest at t = 0.

0%
$54.4\: cm/s$
0%
$44.4\: cm/s$
0%
$64.4\: cm/s$
0%
$14.4\: cm/s$

Explanation

At t = 0 the velocity is zero i.e. the particle is at an extreme The equation for displacement may be written as

$\displaystyle x=A\cos \omega t$
The velocity is

$\displaystyle v=-A\omega \sin \omega t$ At

$\displaystyle t=\frac{1}{6}s$

$\displaystyle v=-(0.1m)(6.28s^{-1})\sin \left ( \frac{6.28}{6} \right )$

$\displaystyle =(-0.628\: m/s)\sin \frac{\pi }{3}=54.4\: cm/s$

A particle is moving on a circular track with uniform speed. Its motion is

0%
Periodic and simple harmonic
0%
Periodic but not simple harmonic
0%
Damped
0%
None of these.

State whether true or false:

The particle is moving simple harmonically

0%
True
0%
False

Explanation

F = 8 - 2x or F = -2(x - 4) at equilibrium position F = 0

$\Rightarrow$ x = 4 is equilibrium position
Hence the motion of particle is SHM with force constant 2 and equilibrium position x = 4
Yes motion is SHM

The equation of displacement of a particle executing simple harmonic motion is x = (5m)

$\displaystyle \sin \left [ (\pi s^{-1})t+\frac{\pi }{3} \right ]$ . Write down the amplitude, time period and maximum speed. Also find the velocity at t = 1s.

0%
$\displaystyle -\frac{5\pi }{2}m/s$
0%
$\displaystyle -\frac{\pi }{2}m/s$
0%
$\displaystyle -\frac{3\pi }{2}m/s$
0%
$\displaystyle -\frac{\pi }{3}m/s$

Explanation

Comparing with equation

$\displaystyle x=A\sin (\omega t+\delta )$ we see that the amplitude = 5 m
and time period =

$\displaystyle \frac{2\pi }{\omega }=\frac{2\pi }{\pi s^{-1}}=2s$
The maximum speed =

$\displaystyle A\omega =5m\times \pi s^{-1}=5\pi \: m/s$
The velocity at time

$\displaystyle t=\frac{dx}{dt}=A\omega \cos (\omega t+\delta )$
At t = 1 s

$\displaystyle v=(5m)(\pi s^{-1})\cos \left ( \pi +\frac{\pi }{3} \right )=-\frac{5\pi }{2}\: m/s$

The left block in figure collides inelastically with the right block and sticks to it. The amplitude of the resulting simple harmonic motion is

0%
$\sqrt{\dfrac{m}{2k}}v$
0%
$\sqrt{\dfrac{m}{k}}v$
0%
$\sqrt{\dfrac{2m}{k}}v$
0%
$\sqrt{\dfrac{m}{4k}}v$

Explanation

Assuming the collision to last for a small interval only we can apply the principle of conservation of momentum The common velocity after the collision is

$\displaystyle \frac{v}{2}$ The kinetic energy =

$\displaystyle \frac{1}{2}(2m)\left ( \frac{v}{2} \right )^{2}=\frac{1}{4}mv^{2}$ This is also the total energy of vibration as the spring is unstretched at this moment If the amplitude is A the total energy can also be written as

$\displaystyle \frac{1}{2}kA^{2}$ Thus

$\displaystyle \frac{1}{2}kA^{2}=\frac{1}{4}mv^{2}$ giving

$\displaystyle A=\sqrt{\frac{m}{2k}}v$

A particle is oscillating simple harmonically with angular frequency

$\omega$ and amplitude

$A$ . It is at a point (A) at certain instant (shown in figure). At this instant it is moving towards mean positive (B). It takes time

$t$ to reach position (B). If time period of oscillations is

$T$ , the average speed between

$A$ and

$B$ is :

0%
$\dfrac {A\sin \omega t}{t}$
0%
$\dfrac {A\cos \omega t}{t}$
0%
$\dfrac {A\sin \omega t}{T}$
0%
$\dfrac {A\cos \omega t}{T}$

Explanation

$x = A\sin \omega t$

$dx = A\omega \cos \omega t \ dt$

$(v) = \dfrac {\int dx}{\int dt}$

$= \dfrac {\int_{0}^{t} A\omega \cos \omega t dt}{\int_{0}^{t} dt}$

$= \dfrac {A\omega \dfrac {\sin \omega t}{\omega}}{t}$

$= \dfrac {A\sin \omega t}{t}$ .

A solid cylinder is attached to a horizontal massless spring as shown in figure. If the cylinder rolls without slipping, the time period of oscillation of the cylinder is?

0%
$2\pi\sqrt{\frac{x}{g}}$
0%
$2\pi\sqrt{\frac{2M}{3K}}$
0%
$2\pi\sqrt{\frac{3M}{8K}}$
0%
$2\pi\sqrt{\frac{3M}{2K}}$

Explanation

Let x be the extension in the spring,
Then,

$x=2R\theta$

$\therefore$ Torque acting on the cylinder about point of contact,

$T=F\times 2R=-Kx(2R)=-4R^2K\cdot \theta$

$\therefore T=2\pi\sqrt{\displaystyle \frac{I}{Ke}}=2\pi\sqrt{\displaystyle\frac{\displaystyle \frac{3}{2}MR^2}{4R^2K}}=2\pi\sqrt{\displaystyle \frac{3M}{8K}}$ .

The time taken by a particle performing S.H.M. to pass from point A to B where its velocities are same is

$2$ seconds. After another

$2$ seconds, it returns to B. Determine the time period of oscillation is (in seconds).

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

2×( time taken to go from a to b + the next time taken to return at b) =

$2\times(2+2)= 8 sec(ans)$

A particle is executing simple harmonic motion along a straight line. At displacements

$\displaystyle { r }_{ 1 }$ and

$\displaystyle { r }_{ 2 }$ from its mean position the velocities are

$\displaystyle { v }_{ 1 }$ and

$\displaystyle { v }_{ 2 }$ . The time period of the particle is:

0%
$\displaystyle { 2\pi \left[ \frac { { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } }{ { r }_{ 2 }^{ 2 }+{ r }_{ 1 }^{ 2 } } \right] }^{ { 1 }/{ 2 } }$
0%
$\displaystyle { \frac { 1 }{ 2\pi } \left[ \frac { { v }_{ 1 }^{ 2 }+{ v }_{ 2 }^{ 2 } }{ { r }_{ 2 }^{ 2 }-{ r }_{ 1 }^{ 2 } } \right] }^{ { 1 }/{ 2 } }$
0%
$\displaystyle { 2\pi \left[ \frac { { r }_{ 2 }^{ 2 }-{ r }_{ 1 }^{ 2 } }{ { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } } \right] }^{ { 1 }/{ 2 } }$
0%
$\displaystyle { 2\pi \left[ \frac { { r }_{ 2 }^{ 2 }-{ r }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 }-{ v }_{ 1 }^{ 2 } } \right] }^{ { 1 }/{ 2 } }$

Explanation

Let the SHM be represented by

$x=Asin\omega t$

Thus

$v=A\omega cos\omega t$

$\implies r_1=Asin\omega t_1$

$v_1=A\omega cos\omega t_1$

$\implies r_1^2+\dfrac{v_1^2}{\omega^2}=A^2$

Similarly,

$r_2^2+\dfrac{v_2^2}{\omega ^2}=A^2$

$\implies r_1^2+\dfrac{v_1^2}{\omega^2}=r_2^2+\dfrac{v_2^2}{\omega^2}$

$\implies \omega=\sqrt{\dfrac{v_1^2-v_2^2}{r_2^2-r_1^2}}$

Thus time period of particle=

$T=\dfrac{2\pi}{\omega}$

$=2\pi[\dfrac{r_2^2-r_1^2}{v_1^2-v_2^2}]^{1/2}$

The spring constants of three springs connected to a mass M are shown in figure. When mass oscillates, what are the effective spring constant and the time period of vibration?

0%
$4k, 2\pi\sqrt{\displaystyle\frac{M}{4k}}$
0%
$3k, 2\pi\sqrt{\displaystyle\frac{M}{3}}k$
0%
$2k, 2\pi\sqrt{\displaystyle\frac{M}{2k}}$
0%
None of these

Explanation

In lower segment/zone, two springs, each of constant k, are in parallel.

$\therefore$ Equivalent spring constant

$=(k+k)=2k$
Now, for the system, there is a spring (of constant

$2k$ ) in upper segment and a spring (of constant

$2k$ ) in lower segment.
They are in series

$\therefore k_r=\displaystyle\frac{k_1k_2}{(k_1+k_2)}=\frac{(2k)(2k)}{(2k+2k)}$

$\therefore k_r=4k^2/4k=k$ and

$\therefore T=2\pi\sqrt{\displaystyle\frac{M}{k}}$ .

A particle of mass

$m$ is located in a one dimensional field where potential energy is given by:

$V(x)=A(1-\cos\ {px})$ where

$A$ and

$p$ are constants.
The period of small oscillations of the particle is

0%
$2\pi\sqrt { \cfrac { m }{ (Ap) } }$
0%
$2\pi \sqrt { \cfrac { m }{ \left( A{ p }^{ 2 } \right) } }$
0%
$2\pi \sqrt { \cfrac { m }{ A } }$
0%
$\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { Ap }{ m } }$

Explanation

We are given that a particle of mass

$m$ is located in a one dimensional potential field and the potential energy is given by:

$V(x)=A(1-\cos {px})$
So, we can find the force experienced by the particle as

$F=-\cfrac{dV}{dx}=-Ap\sin {px}$

For small oscillations, we have,

$F\approx -A{p}^{2}x$

Hence, the acceleration would be given by,

$\alpha=\cfrac{F}{m}=-\cfrac{A{p}^{2}}{m}x$

Also we know that,

$\alpha=\cfrac{F}{m}=-{\omega}^{2}x$

So,

$\omega=\sqrt {\cfrac{A{p}^{2}}{m}}$

$\Rightarrow T=\cfrac{2\pi}{\omega}=2\pi \sqrt {\cfrac{m}{A{p}^{2}}}$

A particle moves with simple harmonic motion in a straight line. In first

$T s$ , after starting from rest it travels a distance

$a$ and in next

$Ts$ , it travels

$2\ a$ in same direction, then

0%
Amplitude of motion is $3\ a$
0%
Time-period of oscillation is $8 t$
0%
Amplitude of motion is $4\ a$
0%
Time-period of oscillation is $6 t$

Explanation

Consider a general equation,

$x=Acos\omega t$

Apply for OP,

$A-a=Acos\omega t$

similarly,

$A-(a+2a)=Acos2\omega t$

[

$\therefore$ time doubled]

from here,

$cos\omega t=\frac { A-a }{ A }$

$cos2\omega t=\frac { A-3a }{ A } \\ cos2\theta =2{ cos }^{ 2 }\theta -1\\ { \left( \frac { A-a }{ A } \right) }^{ 2 }=\frac { A-30+A }{ 2A }$

from here,

$A=2a$

So, amplitude of the motion is

$2a$ .

$2a-a=2acos\omega t\\ \frac { \pi }{ 3 } =\omega t\\ \frac { \pi }{ 3 } =\frac { 2\pi }{ T } t \\ T=6t$

Time period

$=6t$

1026560_768172_ans_861de9ce238849219c259bea567fab28.jpg

To and fro motion is an example of

0%
Oscillatory motion
0%
Periodic motion
0%
Circulatory motion
0%
Both (A) and (B).

As shown in figure a simple harmonic motion oscillator having identical four springs has time period

0%
$\displaystyle T = 2\pi\sqrt\frac{m}{4K}$
0%
$\displaystyle T = 2\pi\sqrt\frac{m}{2K}$
0%
$\displaystyle T = 2\pi\sqrt\frac{m}{K}$
0%
$\displaystyle T = 2\pi\sqrt\frac{m}{8K}$

the system shown in

$Fig. 6.330$ is in equilibrium. Masses

${ m }_{ 1 }$ and

${ m }_{ 2 }$ are

$2 kg$ and

$8 kg$ , respectively. Spring constants

${ k }_{ 1 }$ and

${ k }_{ 1 }$ are

$50 N{ m }^{ -1 }$ and

$70 N{ m }^{ -1 }$ , respectively. If the compression in second spring is

$0.5 m$ . What is the compression in first spring? (Both springs have natural length initially.)

0%
$1.3 m$
0%
$-0.5 m$
0%
$0.5 m$
0%
$0.9 m$

A light uniform rod of Young's modulus Y, cross sectional area A, coefficient of linear expansion

$\alpha$ and length

$l_0$ is rigidly connected to the support at one end and the other end of the rod is connected to the string as shown in the figure. The temperature of the rod is increased by

$\triangle\theta$ with supports remaining fixed. Initially, the spring is in natural length position. Spring force on the rod acts uniformly over the cross section during elongation of the rod. Find the net of elongation of the rod. (Assume thermal strain to be small)

0%
$\frac{Kl_0\alpha\triangle\theta}{(K+YA/l_0)}$
0%
$\frac{YA\alpha\triangle\theta}{(K+YA/l_0)}$
0%
$YA\alpha\triangle\theta$
0%
None of the above

Explanation

When the temperature is increases by

$\triangle \theta$ then will elongation in rod and compression in spring.

At increased temperature the length of rod

$U$ will be

$l= l_{o}+ \triangle l$

$l= l_{o}+ l_{o} \alpha \triangle \theta$

If the rod is elongated by amount

$\triangle x$ then the net compression in the rod

$= l- (l_{o}+\triangle x)$

$=l_{o}+ l_{o} \alpha \triangle \theta - l_{o}- \triangle x$

$= l_{o} \alpha \triangle \theta - \triangle x$

$\Rightarrow$ Strain (in rod)

$=\dfrac{l_{o} \alpha \triangle \theta - \triangle x}{l}$

If the force exerted by rad on spring

$F_{r}$ and force exerted log

$s$ bring is

$F_{s}$ at point

$B$ .

then,

$F_{r}= F_{s}$

Since,

the compression in string

$=$ elongation in rod

$= \triangle x$

$\Rightarrow F_{s}= k \triangle x$

and,

we know that in red the youngs modulus

$y= \dfrac{Stream}{Strain}$

$Y=\dfrac{Fr/A}{(l_{o} \angle \triangle \theta - \triangle x)/l}$

Where,

$A \rightarrow$ cross sectional area of rod

$\Rightarrow F_{r}= \dfrac{AY (l_{o} \alpha \triangle \theta - \triangle x)}{l}$

Since,

$F_{r}= F_{s}$

$\Rightarrow \dfrac{AY(l_{o} \alpha \triangle \theta - \triangle x)}{l}= k \triangle x$

$\Rightarrow AY l_{o} \alpha \triangle \theta - AY \triangle x= k \triangle x l$

$\Rightarrow \triangle x (k l + AY)= AY l_{o} \alpha \triangle \theta$

$\triangle x= \dfrac{AY l_{o} \alpha \triangle \theta }{k (l_{o}+ \triangle l)+ AY}= \dfrac{A Y l_{o} \alpha \triangle \theta }{k l_{o} \left[ \left(2+ \dfrac{\triangle l}{l_{o}} \right)+ \dfrac{AY}{l_{o}} \right]}$

Assuming thermal strain

$\left( \dfrac{\triangle l}{l_{o}} \right)$ to be negligible

$\triangle x=\dfrac{AY \alpha \triangle \theta }{k+ \dfrac{AY}{l_{o}}}$

1443332_878432_ans_94f02a8cfe7c43c1be1ba675a702cfb8.png

A small block is connected to one end of a massless spring of un-stretched length

$4.9$ m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by

$0.2$ m and released from rest at

$t=0$ . It then executes simple harmonic motion with angular frequency

$\omega =\pi /3$ rad/s. Simultaneously at

$t=0$ , a small pebble is projected with speed v form point P at an angle of

$45^o$ as shown in the figure. Point P is at a horizontal distance of

$10m$ from O. If the pebble hits the block at

$t=1$ s, the value of v is (take

$g=10m/s^2$ )

0%
$\sqrt{50}$ m/s
0%
$\sqrt{51}$ m/s
0%
$\sqrt{52}$ m/s
0%
$\sqrt{53}$ m/s

Explanation

The block starts its oscillation from its positive extremum.

$\therefore x_{block}(t)=4.9+0.2\cos \omega t$

$\therefore x_{block}(t=1s)=4.9+0.2\cos \left (\cfrac {\pi}{3}\times 1\right)=5$

$m$

$\therefore$ Range of pebble

$=5m$

So,

$\cfrac {v^2\sin 2\theta}{g}=5$

$\implies v=\sqrt {50}$

$m/s$

A block of mass

$200$ g executing SHM under the influence of a spring of spring constant

$k = 90 N m^{-1}$ and a damping constant

$b = 40 g s^{-1}$ . Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In

$(1/2) = -0.693)$

0%
$7$ s
0%
$9$ s
0%
$4$ s
0%
$11$ s

Explanation

Given data,

mass

$m=200g$

Spring constant

$k=90Nm^{-1}$

Damping constant

$b=40gs^{-1}$

To calculate: Time taken for the amplitude of vibration to drop to half of the initial value

We know that amplitude at any time t can be given as:

$A(t)=A_0e^{-\dfrac{bt}{2m}}$

$T_{1/2}=\dfrac{-0.693l×2×0.2}{40×10^{−3}}=6.93s$

Time taken for its amplitude of vibrations to drop to half of its initial values is

$7s$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 12 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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