MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 5

A particle executing SHM has a maximum speed of

$0.5 m{s}^{-1}$ and maximum acceleration of

$1 m {s}^{-2}$ .

The angular frequency of oscillation is:

0%
$2 rad {s}^{-1}$
0%
$0.5 rad {s}^{-1}$
0%
$2\pi rad {s}^{-1}$
0%
$0.5\pi rad {s}^{-1}$

Explanation

Maximum velocity,

$V_{max} = Aw$

Maximum acceleration,

$a_{max} = Aw^2$

$\implies$ Angular frequency,

$w = \dfrac{a_{max}}{V_{max}} = \dfrac{1}{0.5} =2$

$rad/s$

The black graph pictured below represents the poition -time graph for a spring -mass system oscillating withnsimple harmonic motion.
The colored, dashed graphs represents shapes of possible velocity time graphs for the same motion.
The vertical axis stands for position or velocity, but the scaling does not matter. The time axis is the same for all graphs.
Which colored graph best represents the possible velocity for the mass in this spring

0%
red
0%
blue
0%
green
0%
The velocity graph is the same as the position graph
0%
All three colored graphs are equally possible.

Explanation

For a spring mass system(an SHM), the position of mass is given by

$x=Asin(\omega t+\phi)$

The velocity of the mass would be

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

Hence the functions representing them must be

$90^{\circ}$ out of phase. Only the green curve can represent the velocity-time graph.

The black graph pictured above represents the position-time graph for a spring-mass system oscillating with simple harmonic motion.
The colored, dashed graphs represent shapes of possible velocity time graphs for the same motion.
The vertical axis stands for position or velocity, but the scaling does not matter. The time axis is the same for all graphs.
Which colored graph best represents the possible velocity for the mass in this spring-mass system?

0%
Red
0%
Blue
0%
Green
0%
The velocity graph is the same as the position graph
0%
All three colored graphs are equally possible

Explanation

The given displacement graph of SHM represents the displacement

$y$ as ,

$y=a \sin\omega t$ ,

where

$a=$ amplitude ,

$\omega=$ angular frequency ,

therefore velocity ,

$v=dy/dt=a\omega\cos\omega t$

$v=a\omega\sin(\omega t+\pi/2)$ ,

it is clear that velocity v is ahead of displacement by a phase angle of

$\pi/2$ ,which is represented by green graph of velocity .

In their discussions of SHM text books sometimes consider a thingy attached to a horizontal spring and moving horizontally on a frictionless surface, instead of the hanging thingy that we have been looking at. Suppose that the two springs and the two thingies are identical. Think about whether these two systems are significantly different in other respects and decide which one of the following statements is true.

0%
The systems have different periods because their motions are aligned differently with the gravitational field.
0%
The hanging system has a slightly smaller period because the weight of the spring has to be accounted for.
0%
The hanging system has a slightly larger period because the weight of the spring has to be accounted for.
0%
The two systems have identical periods, no matter what the weight of the spring is.

Explanation

The time period of the

$0\int$ mass system is given by

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

m = mass attached to spring, K = spring constant.

clearly, it is seen that time period only depends on mass attached to spring and spring constant. In the given two cases both are the same no matter what the weight of the spring is.

Which of the following regarding oscillatory motion is true?

0%
Motion of the earth is periodic but not oscillatory because it is not to and fro.
0%
Quivering of the string of the musical instrument is an example of oscillatory motion
0%
Motion of the earth is periodic and oscillatory motion because it is not to and fro.
0%
None of the above

Which of the following is an example of oscillatory motion?

0%
Heart beat of a persion
0%
Motion of earth around the sun
0%
Motion of Hally's comet around the sun
0%
Oscillations of a simple pendulam

An object swinging on the end of a string forms a simple pendulum. Some students (and some texts) often cite the simple pendulum's motion as an example of SHM. That is not quite accurate because the motion is really

0%
approximately SHM only for small amplitudes
0%
exactly SHM only for amplitudes that are smaller than a certain value
0%
approximately SHM for all amplitudes.
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None of the above

Simple harmonic motion (SHM) is a technical term used to describe a certain kind of idealized oscillation. Practically all the oscillations that one can see directly in the natural world are much more complicated than SHM. Why then do physicists make such a big deal out of studying SHM?

0%
It is the only kind of oscillation that can be described mathematically
0%
Any real oscillation can be analysed as a superposition (sum or integral) of SHMs with different frequencies
0%
Physics is concerned mainly with the unnatural world.
0%
Students are too stupid to appreciate the real world.

Which of the following conditions must be satisfied for a body to oscillate or vibrate?
A: The body must have inertia to keep it moving across the mid point of its path.
B: There must be a restoring force to accelerate the body towards the midpoint.
C: The fractional force acting on the body against its motion must be small.

0%
Only A
0%
Only A and B
0%
Only B and C
0%
All A, B and C

The velocity vector

$v$ and displacement vector

$x$ of a particle executing SHM are related as

$\dfrac{v dv}{dx} ={\omega}^2 x$ with the initial condition

$v = v_0$ at

$x = 0$ . The velocity

$v$ , when displacement is

$x$ , is:

0%
$v = \sqrt{v^2_0 + {\omega}^2x^2}$
0%
$v = \sqrt{v^2_0 - {\omega}^2x^2}$
0%
$v = {3}\sqrt{v^3_0 + {\omega}^3x^3}$
0%
$v = v_0 - ({\omega}^3 x^3 e^{x^3})^{1/3}$

Explanation

As it is SHM so the equation of motion will be

$F=-kx$ or

$\dfrac{vdv}{dx}=-\omega^2 x$

Now integrating the expression with boundary condition,

$\int^v_{v_0}vdv=-\omega^2\int^x_0xdx$

$\dfrac{1}{2}[v^2-v_0^2]=-\dfrac{\omega^2x^2}{2}$

$v=\sqrt{v_0^2-\omega^2x^2}$

The displacement of a particle is represented by the equation

$y=3 cos(\dfrac{\pi}{4}- 2 \omega t)$ . The motion of the particle is :

0%
simple harmonic with period $\dfrac{2 \pi}{\omega}$
0%
simple harmonic with period $\dfrac{ \pi}{\omega}$
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periodic but not simple harmonic
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non-periodic.

A particle vibrating simple harmonically has an acceleration of

$16{cms}^{-2}$ when it is at a distance of

$4cm$ from the mean position. Its time period is

0%
$1s$
0%
$2.572s$
0%
$3.142s$
0%
$6.028s$

Explanation

The equation of motion for an SHM is given as

$x=Asin(\omega t+\phi)$

$\implies a=\dfrac{d^2x}{dt^2}$

$=-A\omega^2sin(\omega t+\phi)$

$=-\omega^2 x$

Hence for the given position,

$16=\omega ^2 (4)$

$\implies \omega=2$

Hence the time period of the motion is

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2}=\pi\approx 3.142s$

The amplitude of an oscillating simple pendulum is 10 cm and its time period is 4s. Its speed after is when it passes through its equilibrium position is :

0%
Zero
0%
2.0 m/s
0%
0.3 m/s
0%
0.4 m/s

Explanation

$y = A \sin \omega t$

$\dfrac {dy}{dt} = v= A \omega \cos ( \omega t )$

$= A \times \dfrac {2 \pi}{T} \cos \left( \dfrac {2 \pi}{T} t \right)$

$= \dfrac {2 \pi \times 0.1 }{4} \cos \dfrac { 2 \pi \times 1 }{4}$

$= \dfrac {2 \pi \times 0.1} {4} \cos \dfrac {\pi}{2} = 0$

A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan?
(Take

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

0%
8.0 cm
0%
10.0 cm
0%
Any values less than 12.0 cm
0%
4.0 cm

Explanation

Let the minimum amplitude of SHM is

$a$ .
Restoring force on spring

$\displaystyle F=ka$
Restoring force is balanced by weight

$mg$ of block. For mass to execute simle hormonic motion of amplitude

$a$ ,

$\displaystyle \therefore ka=mg$
or

$\displaystyle a=\frac { mg }{ k }$
Here,

$\displaystyle m=2kg,k=200{ N }/{ m },g=10{ m }/{ { s }^{ 2 } }$

$\displaystyle \therefore \quad a=\frac { 2\times 20 }{ 200 } =\frac { 10 }{ 100 } m$

$\displaystyle =\frac { 10 }{ 100 } \times 100cm=10cm$
Hence, minimum amplitude of the motion should be

$10$ cm, so that the mass gets detached from the pan.

Identify the wrong statement from the following:

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If the length of a spring is halved, the time period of each part becomes $\frac {1}{\sqrt{2}}$ times the original
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The effective spring constant K of springs in parallel Is given by $\frac {1}{K}=\frac {1}{K_1}+\frac {1}{K_2}+ ....$
0%
The time period of a stiffer spring is less than that of a soft spring
0%
The spring constant is inversely proportional to the spring length
0%
The unit of spring constant is $Nm^{-1}$

Explanation

The force exerted by two springs attached in parallel to a wall and a mass exert a force

$k_{eq}=k_{1}x+k_{2}x$

on the mass. Thus, the effective spring constant is given by

$k_{eq}= k_{1}+k_{2}$

A particle moves so that its acceleration a is given by

$a = bn$ , where x is displacement from equilibrium position and b is non negative real constant. The time period of oscillation of the particle is :

0%
$2 \pi \sqrt{b}$
0%
$\dfrac{2 \pi}{\sqrt{b}}$
0%
$\dfrac{2\pi}{b}$
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$2 \sqrt{\dfrac{\pi}{b}}$

Explanation

We have

$a=- bx$
For simple harmonic motion
Or comparing

$\omega^2=b$

$\omega =\sqrt{b}$

$\dfrac{2 \pi}{t}=\sqrt{b}$
or

$t =\dfrac{2 \pi}{\sqrt{b}}$

The simple harmonic motion of a particle is given by

$x=a\sin 2\pi t$ . Then, the location of the particle from its mean position at a time

$1/8$ th of a second is:

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

Explanation

Position of particle executing SHM from mean position

$x=a\sin 2\pi t$

$x=a\sin 2\pi\times \displaystyle\frac{1}{8}$

$x=a\sin \displaystyle\frac{\pi}{4}$

$x=\displaystyle\frac{a}{\sqrt{2}}$ .

A spring with force constant

$k$ is initially stretched by

$x_{1}$ . If it further stretched by

$x_{2}$ , then the increase in its potential energy is

0%
$\dfrac {1}{2}k(x_{2} - x_{1})^{2}$
0%
$\dfrac {1}{2}kx_{2}(x_{2} + 2x_{1})$
0%
$\dfrac {1}{2}\ kx_{1}^{2} + \dfrac {1}{2}\ kx_{2}^{2}$
0%
$\dfrac {1}{2}\ k(x_{1} + x_{2})^{2}$
0%
$\dfrac {1}{2}\ k(x_{1} + x_{2})^{3}$

Explanation

According to the question, in the first condition

$U_{1} = \dfrac {1}{2}kx_{1}^{2} .... (i)$
In the second condition,

$U_{2} = \dfrac {1}{2}k (x_{1} + x_{2})^{2} .... (ii)$
Increment in the potential energy

$= U_{2} - U_{1}$

$= \dfrac {1}{2} k(x_{1}^{2} + x_{2}^{2} + 2x_{1} x_{2}) - \dfrac {1}{2} kx_{1}^{2}$

$= \dfrac {1}{2} k(x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2} - x_{1}^{2})$

$= \dfrac {1}{2} k[2x_{1}x_{2} + x_{2}^{2}]$

$= \dfrac {1}{2} kx_{2} (x_{1} + 2x_{1})$ .

For a simple pendulum, relation between time period

$T$ , length of pendulum

$l$ and acceleration due to gravity

$g$ is given by :

0%
$T=2 \pi \dfrac{l}{\sqrt{g}}$
0%
$T=2 \pi \sqrt{\dfrac{l}{g}}$
0%
$T=2 \pi \dfrac{l}{g}$
0%
None of these

Explanation

Time period of simple pendulum is given by

$T = 2\pi\sqrt{\dfrac{l}{g}}$

where

$l$ is the length of pendulum and

$g$ is the acceleration due to gravity.

A particle which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance

$x$ of the particle from the origin is

$F(x)=-kx+a{ x }^{ 3 }$ . Here,

$k$ and

$a$ are positive constants. For

$x\ge 0$ , the functional form of the potential energy

${U}_{(x)}$ of the particle is :

Explanation

We know

$F=-\cfrac { dU }{ dx }$

$\Rightarrow dU=-F.dx$

$\Rightarrow U=-\int _{ 0 }^{ x }{ \left( -kx+a{ x }^{ 3 } \right) } dx\Rightarrow U=\cfrac { k{ x }^{ 2 } }{ 2 } -\cfrac { a{ x }^{ 4 } }{ 4 } \quad$

$\therefore$ we get

$U=0$ at

$x=0$ and

$x=\sqrt { \cfrac { 2k }{ a } }$

Also we get

$U=$ negative for

$x> \sqrt { \cfrac { 2k }{ a } }$

From the given function we can see that

$F=0$ at

$x=0$ i.e slope of U-x graph is zero at

$x=0$

If the differential equation for a simple harmonic motion is

$\dfrac {d^2y}{dt^2} +2y= 0$ , the time period of the motion is

0%
$\pi \sqrt{2} s$
0%
$\dfrac {\sqrt{2}} {\pi} s$
0%
$\dfrac {\pi}{\sqrt{2}} s$
0%
$2 \pi s$
0%
$\dfrac {\sqrt{\pi}} {2} s$

Explanation

Given,

$\dfrac {d^2y}{dt^2}+2y=0$ ....(i)
But, we know that

$\dfrac {d^2y}{dt^2}+\omega^2y=0$ ....(ii)
Comparing both equation, we get

$\omega^2=2$

$\Rightarrow \omega=\sqrt {2}$
The periodic time

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

$T=\sqrt {2}\pi =\pi \sqrt{2}s$

When a simple pendulum is taken from the Equator to the Pole, its period of oscillation will :

0%
increase
0%
decrease
0%
remains same
0%
can't say

Explanation

Time period of simple pendulum is given by

$T =2\pi\sqrt{\dfrac{l}{g}}$

where

$l$ is the length of the pendulum and

$g$ is the acceleration due to gravity.

We know that

$g_{pole}> g_{equator}$

$\implies \ T_{pole}<T_{equator}$

Thus time period of simple pendulum decreases when it is taken from equator to pole.

Which one of the following is simple harmonic?

0%
Rotation of earth around the sun
0%
Rotation of earth about its own axis
0%
Revolving motion of a top
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Motion of a steel ball in a viscous medium
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Motion of oscillating liquid column in U-tube

The amplitude and the periodic time of a SHM are

$5cm$ and

$6s$ respectively. At a distance of

$2.5cm$ away from the mean position, the phase will be

0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { 5\pi }{ 12 } \quad$

Explanation

The equation of motion is given as

$y=5\sin { \cfrac { 2\pi t }{ 6 } }$
Here,

$y=2.5cm$

$\therefore 2.5=5\sin { \cfrac { 2\pi t }{ 6 } }$

$\Rightarrow \cfrac { \pi }{ 6 } =\cfrac { 2\pi t }{ 6 } \Rightarrow t=\cfrac { 1 }{ 2 } s$

$\therefore$ The phase

$=\cfrac { 2\pi t }{ 6 } =\cfrac { 2\pi }{ 6 } \times \cfrac { 1 }{ 2 } =\cfrac { \pi }{ 6 }$

A 2 kg block is dropped from a height

$0.4 m$ on a spring of force constant

$k=1960 N/m$ .The maximum compression of spring is

0%
$0.1 m$
0%
$0.2 m$
0%
$0.3 m$
0%
$0.4 m$

Explanation

By law of conservation of energy , we have loss in gravitational potential energy=gain in elastic potential energy

$\Rightarrow mg(h+x)=\dfrac{1}{2}kx^{2}$

$2\times 10(0.4+x)=\dfrac{1}{2}\times 1960\times x^{2}$

$\Rightarrow 8+20x=980\,x^{2}$

$\Rightarrow 980x^{2}-20x-8=0$
Solving ,we get

$x=0.1 m$

Four springs having the same force constant and a mass

$M$ are connected between rigid walls as shown in figure. The system with smallest time period is

Explanation

$A)$ Time period

${ T }_{ A }=2\pi \sqrt { \cfrac { m }{ 2k } }$

$B)$ Time period,

${ T }_{ B }=2\pi \sqrt { \cfrac { m }{ k+\cfrac { k }{ 3 } } }$

$C)$ Time period,

${ T }_{ C }=2\pi \sqrt { \cfrac { 4m }{ k } } =4\pi \sqrt { \cfrac { m }{ k } }$

$D)$ Time period,

${ T }_{ D }=2\pi \sqrt { \cfrac { m }{ 4k } } =\pi \sqrt { \cfrac { m }{ k } }$

The displacement of a particle in SHM is

$x=10\sin \begin{pmatrix}2t-\dfrac{\pi}{6}\end{pmatrix}m.$ When its displacement is 6 m, the velocity of the particle (in

$ms^{-1}$ ) is

0%
8
0%
24
0%
16
0%
10
0%
12

Explanation

Given,

$x=10\sin(2t-\pi/6)$

Putting

$x=6 \Rightarrow 6=10\sin(2t-\pi/6)$

So,

$\sin(2t-\pi/6)=3/5$

Using formula,

$\sin^2\theta+cos^2\theta=1$

$cos(2t-\pi/6)=\sqrt{1-(3/5)^2}=4/5$

Now, velocity

$v=\dfrac{dx}{dt}=20cos(2t-\pi/6)=20\times (4/5)=16 \, m/s$

The motion of a particle executing SHM in one dimension is described by

$x=-0.3 \, sin \, \left ( t+\frac {\pi}{4} \right )$ where, x is in metre and t in second. The frequency of oscillation in Hz is

0%
$3$
0%
$\dfrac {1}{2\pi}$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {1}{\pi}$

Explanation

$x=-0.3 \, sin \left ( t+\frac{\pi}{4} \right )$

$x=x_0\, sin (\omega t+\phi )$

$x_0=0.3, \omega =1$ and

$\phi =\frac{\pi}{4}$

$2\pi f=1$

$f=\frac{1}{2\pi}$

Which one of the following represents simple harmonic motion ?

0%
$x^{2}=a+bv$
0%
$x=\sqrt{a+bv^{2}}$
0%
$x=a-bv$
0%
$x=\sqrt{a-bv^{2}}$

Velocity at mean position of a particle executing SHM is v, then velocity of the particle at a distance equal to half of the amplitude is

0%
$\frac {\sqrt{3}}{2}v$
0%
$\frac {2} {\sqrt{3}}v$
0%
$2v$
0%
$v$

Explanation

At mean position

$v = a\omega$
Velocity at half the amplitude is

$v'= \omega \sqrt {a^2-y^2}$

$=\omega \sqrt{a^2-\frac{a^2}{4}}= \frac{\sqrt{3}}{2}a\omega = \frac{\sqrt{3}}{2} v$

The average total energy in one time period of a particle of mass

$m$ executing SHM of amplitude

$a$ and angular velocity

$\omega$ is

0%
$\dfrac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }$
0%
$\dfrac { 1 }{ 4 } m{ \omega }^{ 2 }{ a }^{ 2 }$
0%
$0$
0%
$m{ \omega }^{ 2 }{ a }^{ 2 }$
0%
$\dfrac { 1 }{ 8 } m{ \omega }^{ 2 }{ a }^{ 2 }$

Explanation

The average total energy

$E=\dfrac { 1 }{ 2 } m{ \omega }^{ 2 }{ a }^{ 2 }$
Total energy is not a position function.

A tunnel is dug along the diameter of the earth. A mass

$m$ is dropped into it. How much time does it take to cross the earth?

0%
$169.2\ minutes$
0%
$84.6\ minutes$
0%
$21.2\ minutes$
0%
$42.3\ minutes$

Explanation

The mass will execute SHM in the tunnel.

Time period of oscillation

$T = 2\pi\sqrt{\dfrac{R}{g}}$

where

$R = 6.4\times 10^6 \ m$ and

$g = 9.8 \ m/s^2$

$\implies \ T = 2\pi\sqrt{\dfrac{6.4\times 10^6}{9.8}} \ s = 5075 \ s$

So we get

$T = \dfrac{5075}{60} = 84.6 \ minutes$

Time taken to cross the tunnel

$t = \dfrac{T}{2} = \dfrac{84.6}{2} = 42.3 \ minutes$

One end of a rod of length L is fixed to a point on the circumference of a wheel of radius R. The other end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure. The wheel is rotating with a constant angular velocity

$\omega$ about O. Taking

$\displaystyle T=\frac{2\pi}{P\omega}$ the motion of the rod is?

0%
Simple harmonic with a period of T
0%
Simple harmonic with a period of $T/2$
0%
Not simple harmonic but periodic with a period of T
0%
Not simple harmonic but periodic with a period of $T/2$

Explanation

In simple harmonic motion, oscillations occur about an equilibrium position where net force is zero. Also, at extreme positions, the object is stationary. Hence, the motion is not SHM.

The rod moves periodically because after each time period

$T$ , rod returns in the original position.

A pendulum swings backwards and forwards passing through Y, middle point of the oscillation. The first time of pendulum passes through Y, a stopwatch is started. The twenty-first time the pendulum pass through Y, the stopwatch is stopped. The reading is T.
What is the time period of pendulum?

0%
T/40
0%
T/21
0%
T/20
0%
T/10

Explanation

The time taken to complete one oscillation is called the time period of a pendulum.
Here the time period is the time taken to move from Y to B then back to A and then to Y.
The reading of stopwatch = T
Number of oscillations = 10

$\therefore$ Time period =

$\frac{Time \ Taken}{Number \ of \ oscillations}$
= $$\frac{T}{10}

The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and

${E}_{1}$ and

${E}_{2}$ are the total mechanical energies respectively. Then

0%
${ E }_{ 1 }=\sqrt { 2 } { E }_{ 2 }$
0%
${ E }_{ 1 }=2{ E }_{ 2 }$
0%
${ E }_{ 1 }=4{ E }_{ 2 }$
0%
${ E }_{ 1 }=16{ E }_{ 2 }$

Explanation

For, a S.H.M the total mechanical energy

$E\propto { A }^{ 2 }$ (

$A$ =amplitude)

In case of small circle amplitude

$a$ and energy

${ E }_{ 2 }$ and in case of large circle amplitude is

$2a$ and energy

${ E }_{ 1 }$ .

So,

$\cfrac { { E }_{ 2 } }{ { E }_{ 1 } } =\cfrac { { (a) }^{ 2 } }{ { (2a) }^{ 2 } }$

$\Rightarrow { E }_{ 1 }=4{ E }_{ 2 }$

$*$ N.B: In case of circular motion the radius of circle donates the amplitude of S.H.M when it is projected along its diameter.

A particle performing S.H.M. having amplitude 'a' possesses velocity

$\dfrac {(3)^{1/2}}{2}$ times the velocity at the mean position. The displacement of the particle shall be

0%
$a/2$
0%
$(3)^{1/2}\frac{a}{2}$
0%
$(\dfrac {a}{ {2}})^{1/2}$
0%
$(2)^{1/2}a$

Explanation

In S.H.M the velocity of particle is

given by ,

$\boxed{v^2=w^2(a^2-x^2)}$

Where, a-amplitude of SHM

x-displacement of particle

At mean position

$(x=0)$ , let the velocity

of particle be

$v$ .

$\Rightarrow v^2=w^2a^2$

$\Rightarrow \boxed{w^2=\dfrac{v^2}{a^2}}$ .....(1)

Now, let at a displacement

$x$ , the velocity

of particle is

$\dfrac{\sqrt{3}}{2}v.$ We have to find

$x$ .

$\left(\dfrac{\sqrt{3}}{2}v\right)^2=w^2\left(a^2-x^2\right)$

$\Rightarrow (\dfrac{\sqrt{3}}{2}v)^2=\dfrac{v^2}{a^2}(a^2-x^2)$ from (1)

$\Rightarrow \dfrac{3}{2}v^2=\dfrac{v^2}{a^2}(a^2-x^2)$

$\Rightarrow 3a^2=4(a^2-x^2)$

$\Rightarrow 3a^2=4a^2-4x^2$

$\Rightarrow 4x^2=a^2\Rightarrow x^2=\dfrac{a^2}{4}$

$\Rightarrow \boxed{x=\dfrac{a}{2}}$

An oscillator consists of a block attached to a spring (k = 400 N/m). At some time t, the position (measured from the system's equilibrium location), velocity and acceleration of the block are x = 0.100m, v = 13.6 m/s, and a = 123 m/s

$^2$ . The amplitude of the motion and the mass of the block are

0%
$0.2 m, 0.845 kg$
0%
$0.3 m, 0.765 kg$
0%
$0.4 m, 0.325 kg$
0%
$0.5 m, 0.445 kg$

A particle performs S.H.M of amplitude

$A$ along a straight line. When it is at distance

$\cfrac { \sqrt { 3 } }{ 2 } A$ from mean position, its kinetic energy gets increased by an amount

$\cfrac { 1 }{ 2 } m{ \omega }^{ 2 }{ A }^{ 2 }$ due to an impulsive force. Then its new amplitude becomes

0%
$\cfrac { \sqrt { 5 } }{ 2 } A$
0%
$\cfrac { \sqrt { 3 } }{ 2 } A$
0%
$\sqrt { 2 } A$
0%
$\sqrt { 5 } A$

Explanation

Given,

A S.H.M have amplitude A.

So, its total Initial Kinetic Energy of the system is $\dfrac{1}{2}m{{\omega }^{2}}A$

Energy given to the system is $\dfrac{1}{2}m{{\omega }^{2}}A$ . This energy gets added to system energy.

Total final Kinetic energy is = Total initial kinetic energy + added kinetic energy

$\dfrac{1}{2}m{{\omega }^{2}}A{{'}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}+\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}$

$A'=A\sqrt{2}$

Hence, New amplitude is $A\sqrt{2}$

An oscillator is producing FM waves of frequency 2kHz with avariation of 10kHz. The modulating index=?

0%
0.20
0%
5.0
0%
0.67
0%
1.5

Explanation

Modulation Index ,

$m_f=\dfrac{\text{Maximum frequency deviation}}{\text{Modulating frequency}}$

$\Rightarrow m_f= \dfrac{10 kHz}{2 kHz}$

$\Rightarrow m_f=5$

Three masses of

$500g,300g$ and

$100g$ are suspended at the end of a spring as shown and are in equilibrium. When the

$500g$ mass is removed suddenly, the system oscillates with a period of

$2$ second. When the

$300g$ mass is also removed, it will oscillate with a period

0%
$2$ sec
0%
$4$ sec
0%
$8$ sec
0%
$1$ sec

Explanation

After removing

$500g$ -

$T=2\pi\sqrt{\cfrac{m}{K}}=2sec\\ \quad=2\pi\sqrt{\cfrac{\cfrac{400}{100}}{K}}=2sec\rightarrow (1)$

Now after removing

$300$ g also

$\Rightarrow T=2pi\sqrt{\cfrac{\cfrac{100}{1000}}{K}}\rightarrow (2)$

From equation

$1$ and

$2$

$\cfrac{2}{T}=\cfrac{\sqrt{400}}{\sqrt{100}}\\ \cfrac{2}{T}=\cfrac{20}{10}\\T=1sec$

1188338_874198_ans_11b36ac1cf974c418bf1228f3e2f6370.png

Uniform circular motion can also be represented by a simple harmonic oscillator.
State whether given statement is True/False?

0%
True
0%
False

In an electronic watch, the component corresponding to the pendulum of a pendulum clock is a__?

0%
Diode
0%
Transistor
0%
Crystal oscillator
0%
Balance wheel

In case of a simple harmonic motion, if the velocity is plotted along the X-axis and the displacement (from the equilibrium position)(is plotted along the Y-axis, the resultant curve happens to be an ellipse with the ratio:

$\displaystyle\frac{major axis(along X)}{minor axis (along Y)}=20\pi$
What is the frequency of the simple harmonic motion?

0%
$100$ Hz
0%
$20$ Hz
0%
$10$ Hz
0%
$\displaystyle\frac{1}{10}$ Hz

Explanation

Relation between velocity & x is SHM in H.

$\displaystyle\frac{v^2}{\omega^2A^2}+\frac{x^2}{A^2}=1$

$\rightarrow$ Ellipse
Major axis

$=2\Omega A$
Minor axis

$=2a$
Given:

$\displaystyle\frac{2\Omega a}{2a}$

$\Omega =20\pi$

$2\pi f=20\pi$

$f=10$ Hz.

Which one of the following will take place when a watch based on oscillating spring is taken to a deep mine?

0%
It will indicate the same time on earth
0%
It will become fast
0%
It will become slow
0%
It will stop working

A particle executing simple harmonic motion with time period T. The time period with which its kinetic energy oscillates is

0%
T
0%
$2$ T
0%
$4$ T
0%
$\frac{T}{2}$

Explanation

Kinetic energy of the particle executing simple harmonic motion is periodic with period

$\frac{T}{2}$ .

The time period of simple harmonic motion depends upon the

0%
amplitude.
0%
energy
0%
phase constant
0%
mass

Explanation

The time period of simple harmonic motion depends upon the mass.

For example:- Simple pendulum,

The time period of the simple pendulum is given by

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

From the above equation we can conclude that,

$T\propto \sqrt{m}$ (mass of the particle)

$T\propto \dfrac{1}{\sqrt{k}}$

$(k=force\, constant)$

The correct option is D.

The displacement of a particle varies with time according to the relation

$y = a sin \omega t + b cos \omega t.$

0%
The motion is oscillatory but not SHM.
0%
The motion is SHM with amplitude a + b.
0%
The motion is SHM with amplitude $a^2 + b^2$
0%
The motion is SHM with amplitude $\sqrt{a^2 + b^2}$

Explanation

Given

$x = asin \omega t + b cos \omega t$ .......(i)

Let

$a = A cos \phi$ ........(ii)

and

$b A sin \phi$ ..........(iii)

Squaring adding (ii) and (iii), we get

$a^2 + b^2 = A^2 cos^2 \phi + A^2 sin^2 \phi = A^2$

Eq.(i) can be written as

$x = A cos \phi sin \omega t + A sin \phi cos \omega t = A sin (\omega t + \phi)$

It is equation of SHM with amplitude

$A = \sqrt{a^2 + b^2}$

A block of mass

$m$ is hanging vertically by spring of spring constant

$k$ . If the mass is made to oscillate vertically, its total energy is:

0%
Maximum at the extreme position
0%
Maximum at the mean position
0%
Minimum at the mean position
0%
Same at all positions

A particle executing simple harmonic motion with an amplitude

$5$ cm and a time period

$0.2$ s. The velocity and acceleration of the particle when the displacement is

$5$ cm is

0%
$0.5 \pi m s^{-1}, 0 m s^{-2}$
0%
$0.5 m s^{-1}, -5 \pi^2 m s^{-2}$
0%
$0 m s^{-1}, -5 \pi^2 m s^{-2}$
0%
$0.5\pi m s^{-1}, -0.5 \pi^2 m s^{-2}$

Explanation

A particle executing simple harmonic motion with an amplitude

$=5cm=\cfrac{5}{100}=0.05m$

Time period

$=0.2s$

The velocity and acceleration of the particle when the displacement

$=5cm$

$\omega=\cfrac{2\pi}{T}\\ \quad=\cfrac{2\pi}{0.2}=\cfrac{2\pi}{2}\times10=10\pi rad/s$

And the displacement in

$y$ , then acceleration,

$A=\omega^2y$

And velocity

$v=\omega\sqrt{r^2-y^2}$

$v=10\pi\sqrt{(0.05)^2-(0.05)^2}=0$

When

$y=0, A=(10\pi)^2\times0=0$

$v=10\pi\times\sqrt{(0.05)^2-0^2}=10\pi\times0.05\\ \quad=0.5\pi m/s$

Match the Column I with Column II

0%
A - p, B - q, C - s. D - r
0%
A - s, B - r, C - p, D - q
0%
A - r, B - p, C - s. D - q
0%
A - p, B - r, C - q, D - s

Explanation

Time period of spring block system is given by

$T=2\pi \sqrt { \cfrac { m }{ K{ e }_{ ff } } }$ .....(1)

$K{ e }_{ ff }=$ effective

In series

$K{ e }_{ ff }=\cfrac { 1 }{ { K }_{ 1 } } +\cfrac { 1 }{ { K }_{ 2 } } =\cfrac { { K }_{ 1 }+{ K }_{ 2 } }{ { K }_{ 1 }{ K }_{ 2 } }$ ....(2)

In parallel

$K{ e }_{ ff }={ K }_{ 1 }+{ K }_{ 2 }$ .......(3)

In Diagram (A) and (B) spring are connected in parallel, whereas in (C) and (D) Spring is connected in series.

using equation 1,2,3

option

$B$ will be correct.

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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