MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 6

A spring balance together with a suspended weight of

$2.5$ kg is dropped from a height of

$30$ metres. The reading on the spring balance, while falling, will show a weight of.

0%
$2.5$ kg
0%
$1.25$ kg
0%
$0$ kg
0%
$25$ kg

The equation of a progressive wave can be given by

$y = 15 \sin (660 \pi t- 0.02 \pi x) cm$ . The frequency of the wave is :

0%
$330 Hz$
0%
$342 Hz$
0%
$365 Hz$
0%
$660 Hz$

Explanation

$y = 15$ sin

$(660 \pi t - 0.02 \pi x)$
The general equation of a progressive wave is

$y(x,t)= A \sin (\omega t - kx)$

$\omega=2 \pi \nu$

$\nu= \dfrac{660 \pi} {2 \pi}$

=330 Hz

If a simple harmonic motion is represented by

$\dfrac{d^2 x}{d t^2} + \alpha x = 0$ , its time period is then

0%
$2 \pi \sqrt{\alpha}$ .
0%
$2 \pi \alpha$
0%
$\dfrac{2 \pi}{\sqrt{\alpha}}$
0%
$\dfrac{2 \pi}{\alpha}$

Explanation

The equation of SHM,

$\dfrac{d^2x}{dt^2} + \alpha x = 0$ or

$\dfrac{d^2 x}{dt^2} = - \alpha x$
Comparing it with the equation of SHM

$\dfrac{d^2x}{dt^2} = -\omega^2 x ; \omega^2 = \alpha or \omega = \sqrt{\alpha} \therefore T = \dfrac{2 \pi}{\omega} = \dfrac {2 \pi}{\sqrt{\alpha}}$ .

Motion of an oscillating liquid in a U tube is

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periodic but not simple harmonic
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non - periodic
0%
simple harmonic and time period is independent of the density of the liquid.
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simple harmonic and time period is directly proportional to the density of the liquid.

Explanation

Motion of an oscillating liquid column in a U tube is SHM with period,

$T = 2\pi \sqrt{\frac{h}{g}}$ , where h is the height of liquid column in one arm of U tube in equilibrium position of liquid. Therefore, T is independent of density of liquid.

A Simple harmonic oscillator has a period of $0.01{\text{ }}s$ and an amplitude of $0.2\,m.$ The magnitude of the velocity in m $se{c^{ - 1}}$ at the mean position will be

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$20\,\pi$
0%
$100$
0%
$40\,\pi$
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$100\,\pi$

Explanation

Let equation of SHM is

$x= Asin(\omega t)$ , Where

$A= 0.2$

$m$ ,

$\therefore v= A\omega cos(\omega t)$ ,

$\therefore$ Velocity at mean position

$v= A\omega$

Given, Time Period

$=0.01= \dfrac{2\pi}{\omega}$

$\Rightarrow \omega= 200\pi$

$\therefore$ Velocity

$= 0.2 \times 200 \pi= 40\pi$

The amplitude of a particle in SHM is ${\text{5 }}cms$ and its time period is $\pi .$ At a displacement of $3\;cm$ from its mean position the velocity in cm/sec will be:

0%
8
0%
12
0%
2
0%
16

Explanation

Let equation of SHM is

$x= Asin(\omega t)$ , Where

$A= 5$

$cms$ ,

$\therefore v= A\omega cos(\omega t)$

Given, Time Period

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

$\Rightarrow \omega= 2$

$x= 5 sin (2t)$ ,

At a displacement

$3$

$cm$ ,

$\Rightarrow 3= 5 sin(2t)$

$\Rightarrow cos(2t) = \dfrac{4}{5}$ ,

$v= 5\times 2cos(2t)= 5\times 2\times \dfrac{4}{5}=8 cms^{-1}$

The force on a particle of mass

$10\ g$ is

$(10\hat{i}+5\hat{j})N$ . If it starts from rest, what would be its position at time

$t = 5 s$ ?

0%
$(12500\hat{i}+6250\hat{j})m$
0%
$(6250\hat{i}+12500\hat{j})m$
0%
$(12500\hat{i}+12500\hat{j})m$
0%
$(6250\hat{i}+6250\hat{j})m$

Explanation

Mass of particle

$=10g=10\times { 10 }^{ -3 }Kg$

$F=10\hat { i } +5\hat { j }$

$Ma=10\hat { i } +5\hat { j }$

$\dfrac { 10 }{ 1000 } a=10\hat { i } +5\hat { j }$

$\Rightarrow a=1000\hat { i } +500\hat { j }$

$\Rightarrow \dfrac { dV }{ dt } =1000\hat { i } +500\hat { j }$

$\Rightarrow dV=1000dt\hat { i } +500dt\hat { j }$

$\Rightarrow V=1000t\hat { i } +500t\hat { j }$

$\Rightarrow \dfrac { dx }{ dt } =100t$ and

$\dfrac { dy }{ dt } =500t$

$\Rightarrow \int { dx } =\int _{ 0 }^{ 5 }{ 100t } dt\quad \quad \int { dy } =\int _{ 0 }^{ 5 }{ 500t } dt$

$x=\dfrac { 100{ \left( 5 \right) }^{ 2 } }{ 2 } \hat { i } \quad \quad y=\dfrac { 500{ \left( 5 \right) }^{ 2 } }{ 2 } \hat { j }$

$x=12500\hat { i } \quad \quad \quad y=6250\hat { j }$

position

$=12500\hat { i } +6250\hat { j }$

So, the correct answer is 'A'.

Two particle P and Q are executing simple harmonic motion with equal amplitude and frequency. At a distance of half the amplitude one of them is moving away from the mean position while the other is observed at a distance half the amplitude to be moving towards mean position as shown in figure. The phase difference between them is:

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$\dfrac{\pi}{2}$
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$\dfrac{\pi}{3}$
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$\dfrac{4\pi}{3}$
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$\dfrac{\pi}{6}$

Explanation

Equation of SHM:

$x=a\sin wt$

for

$P$

$a\sin (wt +\phi)=a/2$

$\sin (wt) 1/2$

let

$t=0$

$a\sin \phi=a/2$

$\phi=\pi/6$

for

$Q:a\sin (wt+\delta )=-a/2$

$\sin (\delta )=-1/2$

$\delta =-\pi /6$

Phase difference

$=\phi-\delta$

$=\dfrac{\pi}{6}-\left(-\dfrac{\pi}{6}\right)$

$\boxed{\pi/3}$

1439010_992151_ans_ac2847ffca0b439492f0139081a0316e.png

Figure shows three systems in which a block of mass m can execute S.H.M. What is ratio of frequency of oscillation?

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2: 1: 4
0%
1: 2: 4
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4: 2: 1
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3: 2: 1

Explanation

(i)

$\omega_1=\sqrt{\dfrac{K}{m}}$
(ii)

$k\dfrac{X}{2}=2T$

$T= ma = -\dfrac{kx}{4}$

$\omega_2 =\sqrt{\dfrac{k}{4m}}$

$-4kx = ma$

$a=-\dfrac{4k}{m}x, \omega_3$

$=\sqrt{\dfrac{4k}{m}}$
1124138_993381_ans_2301ec84cd984f7ba5354ab5de5380c9.png

Two blocks A and B of masses m and 2 m, respectively, are held at rest such that the spring is in natural length. Find out the accelerations of both the blocks just after release.

0%
$g\downarrow ,g\downarrow$
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$\dfrac { g }{ 3 } \downarrow ,\dfrac { g }{ 3 } \uparrow$
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$0, 0$
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$g\downarrow ,c$

Explanation

As spring is in natural length then

$a = \left(\dfrac{m_1 + m_1}{m_2 + m} \right) g$

Case (1)

take,

$m_2 = 2m$

$m_1 = m$ then

$a = \left(\dfrac{2m - m}{3m} \right) g$

$a = g/3 \uparrow$

Case (2)

take

$m_2 = m$

$m_1 = 2m$ then

$a = \left(\dfrac{m - 2m}{3m} \right) g$

$a = -9/3$ or

$a = 9/3 \downarrow$

1054869_981471_ans_e01759de860844c28b72b39b81311242.png

A particle executes SHM with an amplitude of

$2cm$ . When the particle is at

$1cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

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$\cfrac { 1 }{ 2\pi \sqrt { 3 } }$
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$2\pi \sqrt { 3 }$
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$\cfrac { 2\pi }{ \sqrt { 3 } }$
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$\cfrac { \sqrt { 3 } }{ 2\pi }$

Explanation

Given,

$A=2cm$

$x=1cm$

The relation of velocity and displacement is given by

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

Acceleration,

$a=-\omega ^2x$

According to question,

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

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

$\dfrac{2\pi}{T}\times 1=\sqrt{2^2-1^2}=\sqrt{4-1}$

$T=\dfrac{2\pi}{\sqrt{3}}$

The correct option is C.

The maximum velocity of a particle executing simple harmonic motion is

$v$ . If the amplitude is doubled and the time period of oscillation decreased to

$1/3$ of its original value, the maximum velocity becomes:

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$18v$
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$12v$
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$6v$
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$3v$

Explanation

Lets consider

$v=$ the maximum velocity of a particle executing harmonic motion

$v=A\omega$ . . . . . . . .(1)

After the time decreased

$1/3$ of its original value,

$T'=\dfrac{1}{3}T$

$A'=2A$

$\omega '=\dfrac{2\pi}{T'}=3\omega$

$v '=A'\omega '$

$v'=2A\times 3\omega$

$v'=6A\omega =6v$ (from equation 1)

The correct option is C.

A particle executes simple harmonic motion between

$x=-A$ and

$x=+A$ . The time taken for it to go from

$0$ to

$A/2$ is

$T_1$ and to go from

$A/2$ to A is

$T_2$ . Then.

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$T_1 < T_2$
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$T_1 > T_2$
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$T_1 = T_2$
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$T_1 =2T_2$

Explanation

Using

$x=A \sin \omega t$

For

$x=A/2,\ sin \omega T_1=1/2 \Rightarrow T_1=\dfrac{\pi}{6 \omega}$

For

$x=A,\ \sin \omega (T_1+T_2)=1 \Rightarrow T_1+T_2=\dfrac{\pi}{2 \omega}$

$\Rightarrow T_2=\dfrac{\pi}{2 \omega}-T_1=\dfrac{\pi}{2 \omega} - \dfrac{\pi}{6 \omega}=\dfrac{\pi}{3 \omega} \ i.e., T_1 < T_2$

Alternative method : In SHM, velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from

$0$ to

$\dfrac{A}{2}$ will be less than the time taken to go from

$\dfrac{A}{2}$ to

$A$ . Hence

$T_1 < T_2$

Two bodies M and N of equal mass are suspended from two separate massless spring of force constant

$k_1$ and

$k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is?

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$\dfrac{k_1}{k_2}$
0%
$\sqrt{\dfrac{k_1}{k_2}}$
0%
$\dfrac{k_2}{k_1}$
0%
$\sqrt{\dfrac{k_2}{k_1}}$

Explanation

$V_{max}=A\omega$

$\implies {V_{max}}_1={V_{max}}_{2}$

$\implies A_1\omega _1=A_2\omega_2$

$\implies \cfrac {A_1}{A_2}=\cfrac {\omega_2}{\omega_1}$

Now,

$K_1=m\omega_1^2$

$\implies \omega_1=\sqrt {\cfrac {K_1}{m}}$

and

$K_2=m\omega_2^2$

$\implies \omega_2=\sqrt {\cfrac {K_2}{m}}$

$\therefore \cfrac {A_1}{A_2}=\sqrt {\cfrac {K_2}{K_1}}$

"The motion of a particle with a restoring force gives oscillatory motion". Which of the following force can be a restoring force for the oscillatory motion.

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Frictional force
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A constant force F
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A time varying force opposite to direction of motion
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Force due to a spring

"A bird flapping its wings circles around a clock tower". Which part of the motion is periodic and oscillatory ?

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Bird's motion is oscillatory while wing flapping was periodic
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Both bird's motion and wing flapping were oscillatory
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Bird's motion is periodic while wing flapping was oscillatory
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Both bird's motion and wing flapping are non-oscillatory

The bye-bye gesture, we do using hands is an example of

0%
Periodic motion
0%
Oscillatory motion
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Linear motion
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Circular motion

Simple harmonic motion of a particle can be described as the motion of a particle in which

0%
smaller oscillations takes place about the mean position
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acceleration of the particle is proportional to displacement
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velocity of the particle is proportional to instantaneous displacement
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displacement of the particle is constant

A particle executes simple harmonic motion with a frequency

$'f'$ . The frequency with which its kinetic energy oscillates is?

0%
$\dfrac{f}{2}$
0%
$f$
0%
$2f$
0%
$4f$

Explanation

Let,

$x=A\sin\omega t$

$v=\cfrac {dx}{dt}=A\omega \cos \omega t$

Kinetic energy,

$K=\cfrac {1}{2}mv^2$

$\implies K=\cfrac {1}{2}m\omega ^2A^2\cos^2\omega t$

$\implies K=\cfrac {1}{2}m\omega^2A^2\left (\cfrac {1+cos2\omega t}{2}\right)$

$\implies K=\cfrac {1}{4}m\omega^2A^2(1+\cos^2\omega t)$

$\therefore \omega_K=2\omega$

$\therefore$ Frequency of oscillation of K.E

$=2f\quad \left[f=\cfrac {\omega}{2\pi}\right]$

Two masses

$m_1$ and

$m_2$ are suspended together by a massless spring of spring constant

$k$ as shown in the figure. When the masses are in equilibrium,

$m_1$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of

$m_2$ .

0%
$\sqrt{\dfrac{k}{3m_2}}$ .
0%
$\sqrt{\dfrac{k}{m_2}}$ .
0%
$\sqrt{\dfrac{k}{2m_2}}$ .
0%
$\sqrt{\dfrac{k}4{m_2}}$ .

Explanation

Here, during oscillation

Mass,

$m = m_2$

$\therefore \omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{k}{m_2}}$

Which of the following equations represents a particle performing simple harmonic motion

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x= 3t
0%
x = 4 sin 2t
0%
x = 1/t
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x = 3 log 2t

In periodic motion, the displacement is

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directly proportional to the restoring force
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inversely proportional to the restoring force
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independent of restoring force
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independent of any force acting on the particle

A particle performing SHM has a period of 6s and amplitude of 8 cm. The particle starts from the mean position and moves towards the positive extremity. At what time will it have the maximum amplitude

0%
6s
0%
4.5s
0%
1.5 s
0%
3 s

What is the velocity of the particle executing SHM at x = A/2, if the angular frequency = 2 rads/s and amplitude is 5 cm:

0%
$5$ cm/s
0%
$0$ cm/s
0%
$5 \sqrt(3)$ cm/s
0%
$5 \sqrt(2)$ cm/s

Explanation

We know that

$v^2=\omega^2(A^2-x^2).$

Substituting the values of A = 5 cm and

$\omega = 2, x= A/2$ , we get,

$v=\omega \sqrt(3) A/2=5 \sqrt(3)$

The correct option is (c)

The displacement of a particle performing linear S.H.M is given by

$x=6 sin(3 \pi t-5\pi/6)$ metre. Find the time at which the particle reaches the extreme position towards the left:

0%
5/9 secs
0%
4/9 secs
0%
1/9 secs
0%
7/9 secs

Explanation

The amplitude at the extremes are +A and -A respectively. Comparing

$x=6 sin(3 \pi t+5\pi/6)$ with

$x=A sin (\omega t + \phi)$ , we get amplitude A = 6.

For the particle to reach the amplitude, we have

$(3 \pi t-5 \pi/6)= \pi/2 \implies 3 \pi t =5 \pi/6 + 3 \pi/6=4 \pi/3$

Thus, t = 4/9 secs

The correct option is (b)

A particle executes SHM along x axis and is at the mean position at t=What is its velocity at its mean position. The amplitude of SHM is 5 cm and angular frequency is 2 rad/s:

0%
5 cm/s
0%
10 cm/s
0%
7 cm/s
0%
0 cm/s

Explanation

The particle has its maximum velocity at mean position and is given by

$v=A \omega = 10$ cm/s

The correct option is (b)

A particle executes SHM given by the equation

$x = 4 sin (2 \pi t +\pi/4)$ , what will be the velocity of the particle at t = (1/8)th sec;

0%
velocity = 0
0%
velocity = maximum
0%
velocity = minimum
0%
The particle's speed cannot be determined

Explanation

$x = 4 sin (2 \pi t +\pi/4)$
Differentiating x w.r.t. t, we get,

$dx/dt =v = 8 \pi cos (2 \pi t +\pi/4)$
Substituting t = 1/8, we get,

$v=8 \pi cos (2 \pi t +\pi/4)= cos (\pi/2)=0$

At t = 1/8th of a sec, the particle's velocity will be zero . Thus we infer that the particle is at the extreme position

The correct option is (a)

For what phase difference between two SHMs will the amplitude of the resultant SHM be zero

0%
$\pi/2$
0%
$2\pi$
0%
$\pi$
0%
$0$

Explanation

Let two SHM equations are

$x_1= Asin(\omega t)$ and

$x_2= Asin(\omega t+ \delta)$ ,

$\Rightarrow x= x_1+ x_2= Asin(\omega t) +Asin(\omega t+\delta)$

$\Rightarrow x= R sin(\omega t+ \phi)$

$\Rightarrow R= \sqrt{A^2 +A^2 +2A\times Acos(\delta)}$

Since

$R=0$ ,

$\Rightarrow A^2 + A^2 + 2A^2cos(\delta)=0$

$\Rightarrow cos(\delta) = -1$

$\therefore \delta = \pi$ , Required phase difference is

$\pi$

A particle performs SHM given by the equation

$x = A sin \omega t$ . Where is the particle at t = 3T/8,

0%
The particle is at a distance of $A/\sqrt(2)$ moving away from the mean position to its extreme
0%
The particle is at a distance of $A/\sqrt(2)$ moving towards the mean position to its extreme
0%
The particle is at a distance of $A(1-1/\sqrt(2))$ moving away from the mean position to its extreme
0%
None of these

Explanation

From the equation of SHM, it is clear that the particle has started from rest at the mean position.

at t=T/4 =2T/8, the particle would have reached the extreme position
at t = 3T/8, the particle will be returning towards the mean position

Substituting t =3T/8, we get,

$x=A sin [(2 \pi /T)(3T/8)]= A sin (3 \pi/4)=A/\sqrt(2)$

The displacement of the particle from mean position is

$A/\sqrt(2)$

The correct answer is (b)

If the phase difference between two sinusoidal waves is

$\pi/2$ , what is the corresponding path difference

0%
$2\lambda/4$
0%
$\lambda$
0%
$\lambda/4$
0%
$\lambda/2$

Explanation

Path difference and phase difference are related by the formula Path difference =

$(\lambda/2 \pi) \times$ phase difference

$=(\lambda/2 \pi) \times (\pi/2)=\lambda/4$

The correct option is (c)

If two SHMs of different amplitudes are added together, the resultant SHM will be a maximum if the phase difference between them is

0%
$\pi/2$
0%
$\pi/4$
0%
$\pi$
0%
$2\pi$

Explanation

If two SHMs of different amplitudes

$A$ and

$B$ are superimposed with each other, then the amplitude of the resultant SHM is given by

$R=\sqrt(A^2+B^2+2AB cos \delta); \delta$ is the phase difference between the SHMs

The resultant amplitude will be maximum, if the phase difference between them is 0 or

$2\pi$

The correct option is (d)

What type of curve do we get, if

$x^2$ and

$v^2$ are plotted for a particle executing SHM, x and v are the position and velocity of the particle:

0%
Circle
0%
Straight line
0%
Ellipse
0%
Rectangle

Explanation

The equation of SHM, when the particle started at the mean position at t = 0 is given by

$x=A sin (\omega t )$

Differentiating x w.r.t. t, we get

$v = A \omega cos \omega t$

To eliminate

$\theta$ , we can do

$(x/A)^2 + (v/A \omega)=1 \implies$ , the plot will be a circle

The correct option is (c)

A particle executes SHM from its mean position at t=0 with an amplitude A, what will be its velocity at x=A/2, in its forward motion towards the extreme:

0%
$\sqrt(3)A \omega/2$
0%
$\sqrt(3)A \omega/4$
0%
$A \omega/2$
0%
$A \omega$

Explanation

The equation of SHM, when the particle started at the mean position at t = 0 is given by

$x=A sin (\omega t )$

At x=(A/2), the particle reaches it at time t given by

$sin \omega t=0.5. \implies \omega t = \pi/6$ .

Differentiating x w.r.t. t, we get

$v = A \omega cos \omega t.$ . Substituting the value of

$\omega t$ , we get

$v =A \omega cos 30 = \sqrt(3)A \omega/2$

The correct option is (a)

The time period of oscillation of magnet in a vibrating magnetometer is

$1.5$ sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be

0%
$0.75$ sec
0%
$1.5$ sec
0%
$3.0$ sec
0%
$6.0$ sec

Explanation

$\begin{array}{l} \dfrac { { { T_{ 1 } } } }{ { { T_{ 2 } } } } =\sqrt { \dfrac { { { M_{ 2 } } } }{ { { M_{ 1 } } } } } =\sqrt { \dfrac { { { M_{ 1 } } } }{ { \dfrac { 4 }{ { { M_{ 1 } } } } } } } =\dfrac { { { T_{ 1 } } } }{ { { T_{ 2 } } } } =\dfrac { 1 }{ 2 } \\ \Rightarrow { T_{ 2 } }=3 \end{array}$

The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from

$10\ cm$ to

$8\ cm$ in

$40$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is

$1.3$ , the time In which amplitude of this pendulum will reduce from

$10\ cm$ to

$5\ cm$ in carbon dioxide will be close to (ln

$5=1.601, \ln { 2 } 2=0.693$ )

0%
$231\ s$
0%
$208\ s$
0%
$161\ s$
0%
$142\ s$

Explanation

$\textbf{Solution}:$

We know that the amplitude of damped oscillation is

$A=A_0 e^{\dfrac{-bt}{2m}}$

$\implies 8=10e^{(-\dfrac{b_{air} \times 40}{2m})}$

and

$5=10e^{-(\dfrac{b_{co_2} \times t_2}{2m})}$

$\implies \dfrac{In\dfrac{5}{4}}{In2}=\dfrac{b_{air} \times 40}{b_{co_2} \times t_2}$

Therefore ,

$t_2=\dfrac{b_{air}}{b_{co}} \times \dfrac{40 \times In2}{In\dfrac{5}{4}}=161s$

$\textbf{Hence C is the correct option}$

The maximum velocity of a particle, executing simple harmonic motion with an amplitude

$7$ mm, is

$4.4$ m/s. The period of oscillation is :

0%
$100$ s
0%
$0.01$ s
0%
$10$ s
0%
$0.1$ s

Explanation

$v= \omega A =\frac{2\pi}{T}A$
Take

$\pi \approx 22/7$

A particle is executing SHM with amplitude of 4 cm and has a maximum velocity of 10 cm/sec.
(a) At what displacement its velocity is 4 cm/sec?
(b) What is its velocity at displacement 2 cm?

0%
$\pm$ 3.66cm, 8.66 cm
0%
$\pm$ 4.66cm, 8.66 cm
0%
$\pm$ 3.66cm, 9.66 cm
0%
$\pm$ 5.66cm, 10.66 cm

Explanation

(a) In SHM velocity at any position y is given by
v =

$\omega$

$\sqrt{A^{2} - y^{2}}$

$V_{max} = A \omega$ at y = 0
10 = 4

$\omega$

$\omega$ = 5/2 rad/sec.
Now V = 4 cm/sec.at
4 = 5/2

$\sqrt{4^{2} - y^{2}}$
16 -

$y^{2}$ = 2.56

$y^{2}$ = 13.46
y =

$\pm$

$\sqrt{13.46}$ cm =

$\pm$ 3.66cm
(b) At y = 2cm
v = 5/2

$\sqrt{4^{2} - 2^{2}}$ = 5/2 x 2

$\sqrt{3}$ = 5

$\sqrt{3}$ = 8.66 cm

The potential energy of a particle of mass 10g varies as its displacement from its mean position given by

$U = 3x^2 + 3$ , then, the particle performs a

0%
SHM with time period $T = \pi$ s
0%
Linear motion
0%
SHM with time period $T =\dfrac{ \pi}{5\sqrt{6}}$ s
0%
SHM with time period $T = \pi/5$ s

Explanation

The force acting on the system is

$F = -dU/dx = -6x \implies a=-(6/0.01)x=-600x$ . The angular velocity of the particle is

$\omega= \sqrt{600}$

Thus, the motion of the particle is SHM with time period

$T=2 \pi/\sqrt{600}$ or

$T = \pi/5\sqrt{6}$ s

The correct option is (c)

At what position along a straight line will the velocity be zero for a particle executing SHM

0%
Mean position
0%
$x=A/2$
0%
Extremes
0%
$x= - A/2$

Explanation

The velocity and displacement follows an equation given by

$v^2=\omega (A^2-x^2)$ . If v has to be zero, then x=A.

If the particle was at the extreme, the particle will have a zero velocity

Thus the correct option is option(c)

The potential energy of a particle is directly proportional to its linear displacement from its mean position. Then, the particle performs a

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Retarding straight line motion
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Damped SHM
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Linear SHM
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Angular SHM

A particle moves along the

$X-$ axis as

$x=u(t-2)+a(t-2)^{2}$

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the initial velocity of the particle is $u$
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the acceleration of the particle is $\alpha$
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the acceleration of the particle is $2\alpha$
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at $t=2s$ particle is at the origin.

Explanation

We know,

$v=\dfrac{dx}{dt}=u+2a(t-2)$ (1)

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

Here we will discuss all the option,

$\textbf{Case 1}$

Initial velocity means, v at t=0,

i.e ,

$v= u-4a \neq u$

Option

$\textbf A$ is correct.

$\textbf{Case 2}$

It is clear from equation (1) that ,

Acceleration

$=2a$

Option

$\textbf B$ is incorrect and option

$\textbf C$ is correct.

At t=2s,

$x=0$ i.e particle is ar origin.

Option

$\textbf D$ is correct.

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that is at a distance

$\dfrac{2A}{3}$ from equilibrium position. The new amplitude of the motion is:

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$3A$
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$A\sqrt3$
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$\dfrac{7A}{3}$
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$\dfrac{A}{3}\sqrt{41}$

Explanation

$3\omega \sqrt {A^2-\left(\dfrac{2A}{3}\right)^2}=\omega \sqrt{A_1^2-\left(\dfrac{2A}{3}\right)^2}$

$\therefore A_1=\dfrac{7A}{3}$

Two particles

$A$ and

$B$ of equal masses are suspended from two massless springs of spring constants

$k_1$ and

$k_2$ respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of

$A$ and

$B$ is :

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$\sqrt{\dfrac{k_1}{k_2}}$
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$\dfrac{k_2}{k_1}$
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$\sqrt{\dfrac{k_2}{k_1}}$
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$\dfrac{k_1}{k_2}$

In SHM which of the following statement is/are correct?

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Displacement and velocity may be in the same direction.
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Displacement and velocity can never be in the same direction.
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Velocity and acceleration may be in the same direction.
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Displacement and acceleration can never be in the same direction.

Explanation

Let ua suppose a particle in SHM with mean at origin,

In SHM,

During motion from mean to extreme,

velocity is along +ve x,

And displacement is also in +ve x,

And,

During motion from extreme to mean,

velocity is along -ve direction

as well as acceleration is along -ve, direction

And,

By ,

$a=\omega^2 x$

It is clear that acceleration and displacement can't

be in same direction,

Hence Option

$\textbf{A,C,D}$ is the correct answer.

A spring of force constant K is attached with a mass executes with S.H.M has a time period T. It is cut into three equal parts. If the springs are connected in parallel then the time period of the combination for the same mass is

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T
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T/3
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T/9
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2T/3

Explanation

A spring of length

$l$ can be considered as a series combination of spring of length

$\cfrac { l }{ 3 }$ and the equivalent spring constant of the original spring. Now, if

$k'$ is the spring constant of each of the three spring then

$k$ is the spring constant of original spring, then

$\cfrac { 1 }{ k } =\cfrac { 1 }{ k' } +\cfrac { 1 }{ k' } +\cfrac { 1 }{ k' }$

$\Rightarrow k=\cfrac { k' }{ 3 }$

$\Rightarrow k'=3k$

The equivalent spring constant of parallel combination of these spring is

$k''=k'+k'+k'=3k'=9k$

The time period of spring mass system of original spring

$T=2\pi \sqrt { \cfrac { m }{ k } }$ , where,

$m$ =MAss attached to spring

The time period of parallel combination spring mass system,

$T''=2\pi \sqrt { \cfrac { m }{ k'' } } =2\pi \sqrt { \cfrac { m }{ 9k } } =\cfrac { 1 }{ 3 } (2\pi \sqrt { \cfrac { m }{ k } } )$

$T''=\cfrac { T }{ 3 }$

A body executing

$S.H.M.$ along a straight line has a velocity of

$3\ ms^{-1}$ when it is at a distance of

$4\ m$ from its mean position and

$4\ ms^{-1}$ when it is at a distance of

$3\ m$ from its mean position. Its angular frequency and amplitude are:

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$2\ rad\ s^{-1}$ & $5\ m$
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$1\ rad\ s^{-1}$ & $10\ m$
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$2\ rad\ s^{-1}$ & $10\ m$
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$1\ rad\ s^{-1}$ & $5\ m$

Explanation

Velocity

$v=\omega \sqrt{A^2-X^2}$

$3=\omega \sqrt{A^2-4^2}$ ...(1)

$4=\omega\sqrt{A^2-3^2}$ ...(2)

$\dfrac{3}{4} =\dfrac{\sqrt{A^2-4^2}}{\sqrt{A^2-3^2}}$

$\dfrac{9}{16} =\dfrac{A^2-16}{A^2-9}$

$16A^2-9A^2=16^2-9^2$

$7A^2=175$

$A^2=25$

$A=5$

From (1),

$3=\omega \sqrt{5^2-4^2}$

$\omega=1 rad/sec$

A particle executes SHM with an amplitude of 20 cm and time period of 12 sec. The minimum time required for it to move between two points 10 cm on either side of the mean position

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1 s
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2 s
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3 s
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4 s

Explanation

Time taken from complete 1 cycle

$= {t_0} + {t_0} = 2{t_0}$

$T = 2{t_0}$

$\begin{array}{l} x=A\sin \, \, w\, \, { t_{ 0 } } & \\ 10=A\sin \frac { { 2\pi } }{ { 12 } } { t_{ 0 } }, & as\, \, T=\frac { { 2\pi } }{ w } ,w\frac { { 2\pi } }{ T } ,T=12 \\ a=20 & \\ \frac { { 10 } }{ { 20 } } =\sin \left( { \frac { \pi }{ 6 } { t_{ 0 } } } \right) & \\ \frac { 1 }{ 2 } =\sin \left( { \frac { \pi }{ 6 } { t_{ 0 } } } \right) & \\ \sin \left( { \frac { \pi }{ 6 } } \right) =\sin \left( { \frac { \pi }{ 6 } { t_{ 0 } } } \right) & \\ as\, \, \frac { 1 }{ 2 } =\sin \frac { \pi }{ 6 } & \\ { t_{ 0 } }=1 & \\ T=2{ t_{ 0 } }=2\times 1=2\sec . & \end{array}$

1215274_1025191_ans_f74c6921fc014d85a23f1364349035c9.PNG

A small solid cylinder of mass M attached to a horizontal massless spring can roll without slipping along a horizontal surface. find its time period.

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2 $\pi$ $\sqrt{M/2k}$
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$2\pi$ $\sqrt{3M/5k}$
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2 $\pi$ $\sqrt{3M/2k}$
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$\pi$ $\sqrt{3M/2k}$

Explanation

At any instant of rolling the cylinder has rotational and translational kinetic energy.

$K_{rot}$ = 1/2 l

$\omega^{2}$ = 1/2 1/2M

$R^{2}$

$v^{2}$ /

$R^{2}$ = 1/4 M

$v^{2}$ .

$K_{trans}$ = 1/2

$Mv^{2}$

$U_{potential}$ = 1/2

$Kx^{2}$
Total energy E =

$k_{rot}$ +

$K_{trans}$ + U
E = 3/4

$Mv^{2}$ + 1/2

$Kx^{2}$
Here DE/dt = 3/4M(2v) dV/dt

$\div$ 1/2 k2x dx/dt = 0
Or, dv/dt = -2k 3M x
Hence

$\alpha$ =

$-\omega^{2}$ x
Therefore the motion is simple harmonic

$\omega$ = 2k/3M
T = 2

$\pi$

$\sqrt{3M/2k}$

A particle moves along the

$x-$ axis according to the equation

$x = 4 + 3 \sin(2\pi t)$ , here

$x$ is in

$cm$ and

$t$ in second. Select the correct alternative(s)

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The motion of the particle is simple harmonic with mean position at $x = 0$
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The motion of the particle is simple harmonic with mean position at $x = 4\ cm$
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The motion of the particle is simple harmonic with mean position at $x = -4\ cm$
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Amplitude of oscillation is $3\ cm$

Explanation

$x=4+3\sin(27\pi t)$

$x-4=3\sin(27\pi t)$

Put

$x-4=y$

$y=3\sin(27\pi t)$

Amplitude

$= 3cm$

Mean position at y=0,

i.e

$x-4=0$

$x=+4\ cm$

Options

$\textbf{B,D}$ are the correct answers.

The elastic potential energy of a stretched spring is given by

$E=50{x}^{2}$ where

$x$ is the displacement in meter and

$E$ is in joule, then the force constant of the spring is

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$100J{m}^{-1}$
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$100Nm$
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$100J/{m}^{2}$
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$50Nm$

Explanation

Let the force constant of the spring be

$k$ .

Elastic potential energy of a spring is given by

$E$ =

$\dfrac{1}{2}kx^{2}$ . ....(i)

Given , elastic potential energy of stretched spring is

$E$ =

$50$

$x^{2}$ ....(ii)

Comparing equations (i) and (ii) :

$\dfrac{1}{2}kx^{2}$ =

$50$

$x^{2}$

We get

$k$ =

$100$

$\dfrac{J}{m^{2}}$ .

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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