MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 7

Calculate the period of oscillations of block of mass m attached with a set of springs as shown

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2 $\pi$ $\sqrt{m/3k}$
0%
2 $\pi$ $\sqrt{3m/2k}$
0%
2 $\pi$ $\sqrt{m/2k}$
0%
2 $\pi$ $\sqrt{3k/m}$

Explanation

We know,

Time period

$(T)=2\pi\sqrt{\dfrac{m}{k_{eq}}}$

Here,

$k_{eq}=\dfrac{2k}{3}$

Hence

$(T)=\sqrt{\dfrac{3m}{2k}}$

Option

$\textbf B$ is the correct answer.

Two particles A and B are revolving on two circles with time periods 4 second and 6 second respectively. Time period of particle A with respect to B will be

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6 seconds
0%
12 seconds
0%
18 seconds
0%
24 seconds

Explanation

The correct option is B.

We have,

Two particles A and B

$t_1=4s,t_26s$

The time period of A with respect to B is

$2\pi$ times the reciprocal of the relative angular speed of A with respect to B.

Relative angular speed,

$\omega_{AB}=2\pi\left (\dfrac{1}{t_A}-\dfrac{1}{t_B} \right )$

$=2\pi\left ( \dfrac{1}{4s}-\dfrac{1}{6s} \right )$

$=\dfrac{2\pi}{12s}$

Thus, Time period of A with respect to B is:

$\dfrac{2\pi}{\omega_{AB}}=12s$

A particle executes SHM with a time period of 12 s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

0%
6
0%
1
0%
7
0%
8

Explanation

We know,

Equation of SHM,

$y=A\sin\left(\dfrac{2\pi t}{T}\right)$

Where, A=Amplitude,T=time period

$y=0$ mean position,

$0=A\sin\left(\dfrac{\pi t}{6}\right)$

i.e

$t=0$

$y=\dfrac{A}{2}$

$\dfrac{A}{2}=A\sin\left(\dfrac{\pi t}{6}\right)$

i.e

$t=1s$

Difference in times, is the time taken to reach from mean to half of the Amplitude,

i.e

$1\ sec$

Option

$\textbf B$ is the correct answer.

The acceleration displacement

$(a-x)$ graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.

0%
$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{\alpha}}$
0%
$\dfrac{1}{2\pi}\sqrt{\dfrac{\alpha}{\beta}}$
0%
$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{2\alpha}}$
0%
$\dfrac{1}{2\pi}\sqrt{\dfrac{2\beta}{\alpha}}$

Explanation

Equation of given graph will be,

$a=-\dfrac{\beta}{\alpha}X$ ....(1)

$a=-\omega^2X$ ...(2)

Comparing both equations,

$\omega=\sqrt{\dfrac{\beta}{\alpha}}$

So, frequency,

$f = \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{\alpha}}$

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is :

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$m (g + 2\pi^2h)$
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$m (g + \pi^2h)$
0%
$m (g + \frac{\pi^2}{2}h)$
0%
$m (g + \frac{\pi^2}{3}h)$

Explanation

Tesnsion in the string when bob passes through lowest point

$T=mg+\dfrac{mv^2}{r}=mg+mv \omega \ \ (\because v=r \omega)$
putting

$v=\sqrt{2gh}$ and

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

$T=m(g+ \pi \sqrt{2gh})$

A particle of mass

$0.2\ kg$ is executing

$SHM$ . The velocity displacement curve for which is shown in figure. The frequency of oscillation is:

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

A clock which has pendulum made of brass keeps correct time at

$30^\circ {\text{C}}$ . How many seconds it will lose or gain in a day if temperature falls to

$0^\circ {\text{C}}$ .

$\left( {\alpha _{brass} = 1.8 \times 10^{ - 3} /^\circ {\text{C}}^{ - 1} } \right)$

0%
$23.33 sec$
0%
$24.33 sec$
0%
$26.33 sec$
0%
$25.33 sec$

Explanation

$T_0=2\pi\sqrt{\cfrac{L_0}{g}}\\T+2\pi\sqrt{\cfrac{L}{g}}\\T=T_0(1+\cfrac{\alpha\Delta\theta}{2})T=30[1+(1.8\times10^{-3}\times\cfrac{30}{2})]\\T=30[1+(0.9\times10^{-3}\times30)]\\T=23.33sec$

A thin spherical shell of mass

$M$ and radius

$R$ has a small hole. A particle of mass

$m$ is released at its mouth. Then

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the particle will execute SHM inside the shell
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the particle will oscillate inside the shell, but the oscillations are not simple harmonic
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the particle will not oscillate, but the speed of the particle will go on increasing
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none of these

A body of mass

$m$ has time period

$T_1$ with one spring and has time period

$T_2$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period

$T$ is given as

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$T^{2}= T_{1}^{2}+ T_{2}^{2}$
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$T= T_{1}+ T_{2}$
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$\dfrac{1}{T}=\dfrac{1}{ T_{1}}+\dfrac{1}{ T_{2}}$
0%
$\dfrac{1}{ T^{2}}=\dfrac{1}{ T_{1}^{2}}+\dfrac{1}{ T_{2}^{2}}$

Explanation

Angular frequency, $\omega =\sqrt{\dfrac{k}{m}}\ \ \ \Rightarrow \ \dfrac{2\pi }{T}=\sqrt{\dfrac{k}{m}}$

$\Rightarrow \ k=m{{\left( \dfrac{2\pi }{T} \right)}^{2}}$ where, $T$ is time period.

Net Spring constant when two spring is connected in parallel.

$k={{k}_{1}}+{{k}_{2}}$

$\Rightarrow m{{\left( \dfrac{2\pi }{T} \right)}^{2}}=m{{\left( \dfrac{2\pi }{{{T}_{1}}} \right)}^{2}}+m{{\left( \dfrac{2\pi }{{{T}_{2}}} \right)}^{2}}$

$\Rightarrow \dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T}_{1}}^{2}}+\dfrac{1}{{{T}_{2}}^{2}}$

Two simple harmonic motions are represented by the equations

$y_1 = 0.1 sin (100 \pi t + \frac{\pi}{3})$ and

$y_2 = 0.1 cos \pi t$ .
The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is

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$\dfrac{-\pi}{6}$
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$\dfrac{\pi}{3}$
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$\dfrac{-\pi}{3}$
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None of these

The string of a pendulum is horizontal initially. The mass of the bob attached is m. Now the string is released. 'R' is the length of the pendulum.
(i) Velocity of the string at an angle

$30^\circ$ with the vertical is ________

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$\sqrt{\sqrt{2} gR}$
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$\sqrt{\sqrt{3} gR}$
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$\sqrt{\sqrt{5} gR}$
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$\sqrt{\sqrt{7} gR}$

A bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of

$\sqrt{3 gl}.$ Find the angle rotated by the string before it becomes slack.

0%
$\theta = cos^{-1} (-\dfrac{2}{3})$
0%
$\theta = cos^{-1} (-\dfrac{3}{4})$
0%
$\theta = cos^{-1} (-\dfrac{1}{3})$
0%
$\theta = cos^{-1} (-\dfrac{5}{3})$

Explanation

$V = \sqrt {3gl}$

$\dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = - mgh$

${v^2} = {u^2} - 2g\left( {1 + \cos \theta } \right)$

$\Rightarrow {v^2} = 3gl - 2gl\left( {1 + \cos \theta } \right)$ --------------

$(1)$

Again

$,$

$\dfrac{{m{v^2}}}{l} = mg\cos \theta$

${v^2} = \lg \cos \theta$

from equation

$(1)$ and

$(2),$ we get

$3gl - 2gl - 2gl\cos \theta = gl\cos \theta$

$3\cos \theta = 1 \Rightarrow \cos \theta = \frac{1}{3}$

$\theta = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

$,$ angle rotated before the string becomes slack

$= {180^0} - {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$

Hence,

option

$(C)$ is correct answer.

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

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3.14 m/s
0%
6 m/s
0%
1.54 m/s
0%
8.26 m/s

Explanation

P.E of pendulum is converted in to K.E and lost energy

$P.E=K.E+lost energy\rightarrow1$

given lost energy =10% of P.E=0.1P.E

P.E=K.E+0.1P.E

0.9P.E=K.E

$0.9mgh=0.5mv^2$

$v^2=1.8gh$

h=lenght of the pendulum=2

v = speed at lowest point

g=9.8

now

$v=5.94m/s$

The velocity of a particle under going SHM at the mean position is

$2{ms}^{-1}$ . The velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude is

0%
$1.732m/s$
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$\sqrt {3/2}m/s$
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$1/2m/s$
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$\cfrac{1}{\sqrt{3}}m/s$

Explanation

We know that for a particle undergoing SHM with an amplitude 'a', velocity 'v' and angular velocity '

$\omega$ '

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

At mean position x=a

And given v = 2m

${s^{ - 1}}$ at mean position

$\begin{array}{l} 2=\omega \sqrt { { a^{ 2 } } } =\omega a & -1 \\ v=\omega \sqrt { { a^{ 2 } }-\frac { { { a^{ 2 } } } }{ 4 } } =\omega \sqrt { \frac { { 3{ a^{ 2 } } } }{ 4 } } =\frac { { \sqrt { 3 } a\omega } }{ 2 } & -2 \\ Dividing, & \\ \frac { 2 }{ v } =\frac { { \omega a } }{ { \frac { { \omega a\sqrt { 3 } } }{ 2 } } } =\frac { 2 }{ { \sqrt { 3 } } } & \\ v=\sqrt { 3 } m{ s^{ -1 } }=1.732m{ s^{ -1 } } & \end{array}$

The equation of S.H.M of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase

$30^0$ is

0%
$x=4 \, sin (2 \pi t+\pi/6)$
0%
$x=4 \, sin (5 \pi t+\pi/6)$
0%
$x=4 \, sin (2 \pi t+\pi/3)$
0%
$x=4 \, sin (5 \pi t+\pi/3)$

A particle execute S.H.M from the mean position. its amplitude is

$A$ , its time period is "T'. At what displacement,its speed is half of its maximum speed.

0%
$\dfrac{\sqrt{3}A}{2}$
0%
$\dfrac{\sqrt{2}}{3}A$
0%
$\dfrac{2A}{\sqrt{3}}$
0%
$\dfrac{3A}{\sqrt{A}}$

Explanation

For a SHM velocity,

$v=\omega \sqrt{A^2-x^2}$ . . . . . . .(1)

The maximum velocity is at

$x=0$

$v_m=\omega A$

Now,

$v=\dfrac{\omega A}{2}$

From equation (1),

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

$\dfrac{A^2}{4}=A^2-x^2$

$x^2=\dfrac{3A^2}{4}$

$x=\dfrac{\sqrt{3}A}{2}$

The correct option is A.

The maximum velocity of a particle performing linear S.H.M. is

$0.16\ m/s$ and its maximum acceleration is

$0.64\ m/s^{2}$ . The calculate its time period

0%
$1.571\ s$
0%
$0.08\ s$
0%
$16\ s$
0%
$3.2\ s$

Explanation

Given,

$v_{max}=0.16m/s$

$a_{max}=0.64m/s^2$

Time period,

$T=?$

In S.H.M, we know that

$v_{max}=A\omega$ . . . . . . .(1)

$a_{max}=A\omega ^2$ . . . . . . (2)

$\dfrac{a_{max}}{v_{max}}=\omega=\dfrac{2\pi}{T}$

$T=\dfrac{2\pi\times v_{max}}{a_{max}}$

$T=\dfrac{2\times 3.14\times0.16}{0.64}$

$T=1.57s$

The correct option is A.

Two particles are executing simple harmonic motion with same angular frequency but different amplitudes

$A$ and

$2A$ about same mean position on the same straight line. They cross each other when they are at a displacement of

$\dfrac{4A}{5}$ from mean position in same direction. Then phase difference between them is:-

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Zero
0%
$\cos^{-1}(\dfrac{1}{5}) -\cos^{-1}(\dfrac{2}{5})$
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$53^\circ$
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$\cos^{-1}(\dfrac{2}{5}) -\cos^{-1}(\dfrac{4}{5})$

A body executing SHM at a displacement 'x' its PE is

$E_1$ , at a displacement 'Y' its PE is

$E_2$ The P.E at a displacement

$(x+y)$ is

0%
$\sqrt{E}=\sqrt{E_1}-\sqrt{E_2}$
0%
$\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$
0%
$E=E_1+E_2$
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$E=E_1-E_2$

A thin rectangular magnet suspended freely has a period of oscillation

$4\ s$ . If it is broken into two halves each having half their initial length, then when suspended similarly, the time period of oscillation of each part will be

0%
$4\ s$
0%
$2\ s$
0%
$1\ s$
0%
$4\sqrt{2}\ s$

Explanation

A thin rectangular magnet suspended freely has a period of oscillation

$T=4s$
We know that

$T=2\pi\sqrt{\cfrac{l}{g}}\\ \cfrac{T_1}{T_2}=\sqrt{\cfrac{l_1}{l_2}}\\ \cfrac{4}{T_2}=\sqrt{\cfrac{l_1}{l_1/2}}=\sqrt2\quad or\quad T_2=4\sqrt2s$

The vertical extension in a light spring by a weight of

$1\ kg$ suspended from the wire is

$9.8\ cm$ . The period of oscillation:

0%
$20 \pi \,sec$
0%
$2 \pi \,sec$
0%
$\dfrac {2 \pi}{ \ 10}sec$
0%
$200 \pi \,sec$

Explanation

$\because mg=kx\Rightarrow \dfrac mk =\dfrac xg$

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

$T=2\pi \sqrt{\dfrac{9.8\times 10^{-2}}{9.8}}=\dfrac{2\pi }{10}\sec$

A particle execute SHM with time period

$T$ and amplitude

$A$ . The maximum possible average velocity in time

$\dfrac {T}{4}$ :

0%
$\dfrac {2A}{T}$
0%
$\dfrac {4A}{T}$
0%
$\dfrac {8A}{T}$
0%
$\dfrac {4\sqrt {2}A}{T}$

Explanation

Given that,

Amplitude = $A$

Time period = $T$

Average velocity in time period = $\dfrac{T}{4}$

The displacement equation of SHM

$x=A\sin \omega t$

We know that,

$v=\dfrac{dx}{dt}$

$v=\dfrac{d\left( A\sin \omega t \right)}{dt}$

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

Now, the average velocity is

$<v{{>}_{0\to \dfrac{T}{4}}}=\dfrac{\int\limits_{0}^{\dfrac{T}{4}}{vdt}}{\int\limits_{0}^{\dfrac{T}{4}}{dt}}$

$<v>\,=\dfrac{\int\limits_{0}^{\frac{T}{4}}{A\omega \cos \omega t}}{\dfrac{T}{4}}$

$<v>=\dfrac{4A\omega }{T}\int\limits_{0}^{\frac{T}{4}}{\cos \omega tdt}$

$<v>=\dfrac{4A\omega }{T}\left[ \frac{\sin \omega t}{\omega } \right]_{0}^{\dfrac{T}{4}}$

$<v>=\dfrac{4A}{T}\left[ \sin \dfrac{2\pi }{T}\times \dfrac{T}{4}-0 \right]$

$<v>=\dfrac{4A}{T}$

Hence, the average velocity is $\dfrac{4A}{T}$

If y denotes the displacement and t denote the time and the displacement is given by

$y=a\,sin\,\omega t$ , the velocity of the particle is

0%
$a\, cos\,\omega t$
0%
$-a\, cos\,\omega t$
0%
$a \omega\, cos\,\omega t$
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$(a\,cos\,\omega t)/\omega$

For a particle

$SHM$ , equation of motion is given as

$\dfrac{d^{2}x}{dt^{2}}+4x=0$ . The time period of this wave is

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

The displacement of a particle in S.H.M. is indicated by equation

$y\ = \ 10\ sin(20t\ +\ \pi/3)$ where

$y$ is in metres. The value of time period of vibration will be (in seconds):

0%
$10\pi sec$
0%
$\pi /10 sec$
0%
$10 sec$
0%
$200sec$

Two identical bar magnets are placed on above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic filed is T. The time period of each magnet in the same field is

0%
$\sqrt{2}T$
0%
$\sqrt[4]2T$
0%
$\dfrac 1{\sqrt[4]2}T$
0%
$\dfrac 1{\sqrt2}T$

A particle is oscillating according to the equation

$X = 7\cos 0.5\pi t$ , where

$t$ is in seconds. The point moves from the position of equilibrium to maximum displacement in time.

0%
$4.0\ sec$
0%
$2.0\ sec$
0%
$1.0\ sec$
0%
$0.5\ sec$

Explanation

From given equation

$\omega =\dfrac {2\pi}{T}=0.5 \pi \Rightarrow T=4$ sec
Time taken from mean position to the maximum displacement

$=\dfrac {1}{4}T=1$ sec.

The value of phase at maximum displacement from the mean position of a particle in

$S.H.M.$ is :

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

The co-ordinates of a moving particle at a time

$t$ , are given by ,

$x= 5\ sin\ 10t$ ,

$y= 5\ cos\ 10t$ . The speed of the particle is:

0%
$25$
0%
$50$
0%
$10$
0%
None

The displacement of a particle in S.H.M. is indicated by equation

$y\ = \ 10\ sin(20t\ +\ \pi/3)$ where

$y$ is in metres. The value of time period of vibration will be (in seconds):

0%
$10/\pi$
0%
$\pi /10$
0%
$2\pi /10$
0%
$10/\pi 2$

Explanation

$y=10\sin(20t+\dfrac{\pi}{3})$

Comparing with below equation,

$y=A\sin(\omega t+\phi)$ where

$\omega=$ angular velocity=20

Time period

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{\omega}=\dfrac{\pi}{10}$

A body is executing S.H.M. When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocity of the body is 10 cm/sec and 8 cm/sec. Then the time period of the body is:

0%
$2\pi \sec$
0%
$\pi /2\sec$
0%
$\;\pi \sec \;$
0%
$3\pi /2\sec$

Explanation

When the displacement from the mean position is

$4\ cm$

Potential energy=

$\cfrac{1}{2} 4^2u\\=8u\\=8(m\omega^2)$

Kinetic energy

$=\cfrac{1}{2}mv^2\\=\cfrac{1}{2}10^2m\\=50m$

Total energy

$=8m\omega^2+50m$

When the displacement from the mean position is

$5\ cm$

Potential energy=

$\cfrac{1}{2} 5^2u\\=12.5u\\=12.5(m\omega^2)$

Kinetic energy

$=\cfrac{1}{2}mv^2\\=\cfrac{1}{2}8^2m\\320m$

Total energy

$=12.5m\omega^2+32m$

Now as per conservation of energy

$=8m\omega^2+50m$

$=12.5m\omega^2+32m$

$\implies \omega=2$

And

$\omega=\cfrac{2\pi}{T}=2\implies T=\pi\ s$

The displacement of a particle in SHM is indicted by equation

$y=10sin(20t+\dfrac{\pi} {3})$ where y is in metres. The value of the maximum velocity of the particle will be?

0%
$100 m/s$
0%
$150 m/s$
0%
$200 m/s$
0%
$400 m/s$

Explanation

Here,

$Y=A\sin(\omega t+\phi)$

$v_{max}=A\omega$

Comparing,

$v_{max}=10\times 20=200m/s$

Option

$\textbf C$ is the correct answer

A particle is executing simple harmonic motion between extreme positions given by

$(-1.-2.-3)cm$ and

$(1,2,1)cm$ . Its amplitude of oscillation is :

0%
$6cm$
0%
$4cm$
0%
$2cm$
0%
$3cm$

Explanation

Let

$P(-1, -2, -3)\ cm$ and

$Q(1, 2, 1)\ cm$ are the points of extreme positions in S.H.M.

Then

$2 \times A=d(P, Q)$

$=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

$=\sqrt{(1-(-1))^2+(2-(-2))^2+(1-(-3))^2}$

$=\sqrt{2^2+4^2+4^2}$

$2A=\sqrt{36}$

Amplitude

$=\dfrac{6}{2}=3 cm$

1458076_1122257_ans_453a7a1b1f6d4751a145cca2777e9946.png

An object of mass

$m$ is suspended from a spring and it executes

$S.H.M$ with frequency

$v$ . If the mass is increased

$4$ times, the new frequency will be:

0%
$2 v$
0%
$v/2$
0%
$v$
0%
$v/4$

The acceleration of a particle in SHM at 5 cm from the mean position of

$20 cm/sec^2$ . The value of angular frequency in radians per second will be

0%
2
0%
4
0%
10
0%
14

Explanation

$\textbf{Step 1 - Acceleration of particle}$

Acceleration of particle at distance

$x$ from mean position,

$a = -\omega^{2}x$

Given

$a = -20\ cm/\sec^{2}$ [acceleration is towards mean position always so taken negative]

$-20 = -\omega^{2}x$ [here

$\omega$ is angular frequency]

$-20 cm/\sec^{2} = -\omega^{2} \times 5\ cm$

$\omega = 2\ rad/\sec$

Hence option (A) correct.

In case of a simple pendulum, time period versus length is depicted by

Two block of mass

$3\ kg$ and

$6\ kg$ respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant

$k=200\ N/m$ . Initially the spring is unstreatched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be:

0%
$30\ cm$
0%
$25\ cm$
0%
$20\ cm$
0%
$15\ cm$

Explanation

By conserving linear momentum,

$\Rightarrow -3 \times 1+6 \times 2=6V-3V$

or,

$9=9V$

or,

$V=1$

$W_{dome}=\Delta K_E$

$W_{dome}=\cfrac {1}{2}\times 3(1)^2 +\cfrac {1}{2}\times 6(2)^2-\cfrac {1}{2}\times (3)^2$

$\Rightarrow \cfrac {1}{2}(\bar kx^2)=\cfrac {1}{2}[3+24-9]$

$\Rightarrow 200 \times x= 18$

$\Rightarrow x=0.3m=30cm$

The amplitude of a particle in SHM is 5 cm and its time period is

$\pi$ . At displacement of 3 cm from its mean position, the velocity in centimetres per second will be ?

0%
8
0%
12
0%
6
0%
2

The equation of motion of a particle executing SHM is

$(\dfrac{d^2x}{dt^2})+kx=0$ . The time period of the particle will be ?

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

Explanation

The equation of motion,

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

$a=-kx$

$\omega ^2=k$ (

$a=-\omega ^2 x$ )

$\omega =\sqrt{k}$

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

Time period,

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

The correct option is A.

Which one of the following equations of motion represents simple harmonic motion?

0%
Acceleration = $-k_0x+k_1x^2$
0%
Acceleration = $-k(x+a)$
0%
Acceleration = $k(x+a)$
0%
Acceleration = $kx$

A body of mass

$5\times 10^{-3}\ kg$ is making

$S.H.M.$ with amplitude

$1\times10 ^{-1}m$ with maximum velocity

$1ms^{-1}$ , its velocity will be half at displacement of

0%
$5\times 10^{-3}m$
0%
$2.5\times 10^{-3}m$
0%
$5\sqrt{3}\times 10^{-2}m$
0%
$\dfrac{5}{4}\times 10^{-3}m$

Which of the following equations do not represent a simple harmonic motion

0%
$x= A cos (\omega t - \phi) + B sin (\omega t + \phi)$
0%
$x=B tan (\omega t + \phi)$
0%
$x= A e^{i(\omega t - \phi)}$
0%
$x = Ae^{i \omega t} + B e^{i(\omega t -\phi)}$

A particle is performing simple harmonic motion having time period

$3s$ is in phase with another particle which is also undergoing simple harmonic motion at

$t= 0$ . The time period of second particle is T (less than 3s). If they are again in the same phase for the third time after

$45s$ , then the value of T is

0%
$2.8s$
0%
$2.7s$
0%
$2.5s$
0%
$3.2s$

Explanation

Let

$\omega_1$ and

$\omega_2$ be the angular frequencies of first and second particle, respectively. Then, the phase by which they will proceed in time

$t$ is

$\omega_1t$ and

$\omega_2t$ , respectively.
According to the given situation,

$\omega_2t-\omega_1t=3\times 2\pi$ for

$t=45\ s$

$\dfrac{2\pi}{T}-\dfrac{2\pi}{3}=\dfrac{3\times 2\pi}{45}$

$\dfrac{1}{T}=\dfrac{1}{3}+\dfrac{1}{15}=\dfrac{6}{15}\Rightarrow T=2.5\ s$

A body executing SHM has a maximum velocity of

$1\ ms^{-1}$ and a maximum acceleration of

$4ms^{-1}$ . Its time period of oscillation is

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$3.14\ s$
0%
$1.57\ s$
0%
$6.28\ s$
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$0.25\ s$

The time period of oscillation of a particle, that executes SHM, is

$1.2s$ . The time, starting from mean position, at which its velocity will be half of its velocity at mean position is ?

0%
$0.1\ s$
0%
$0.2\ s$
0%
$0.4\ s$
0%
$0.6\ s$

Explanation

Time is zero at extreme position

$T = 1.2\,sec$

$u = \omega \sqrt{A^{2}-x^{2}}; u_{max} = \omega A$

$u = \dfrac{u_{max}}{2} = \dfrac{\omega A}{2} = \omega \sqrt{A^{2}-x^{2}}$

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

$x = A\,cos\,\omega t$

$\dfrac{\sqrt{3}}{2}A = A\,cos\,,\omega t$

$\omega t = \dfrac{x}{6}$

$t = \dfrac{\pi }{6\times \omega }$

$t = \dfrac{\pi }{6}\times \dfrac{1.2}{2\pi } = 0.1\,sec$

1246459_1138746_ans_8d7f48a63d88496c8cf01d050e0677f5.PNG

A particle is executing S.H.M. between

$x = \pm A.$ The time taken to go from 0 to

$\frac {A}{2}$ is

$T_1$ and to go from

$\frac {A}{2}$ 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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$T_1 = T_2$
0%
$T_1 = 2T_2$

Explanation

let

$t=0, x=0$

Then,

$x=A\sin (wt)$

$t=T_1, x=\dfrac{A}{2}$

$\Rightarrow \dfrac{A}{2}=A\sin (wT_1)$

$\Rightarrow \sin wt_1=1/2$

$\Rightarrow T_1=\dfrac{\pi}{6w}$

$t=(T_1+T_2), x=A$

$\Rightarrow A=A\sin (w[T_1+T_2])$

$\Rightarrow \sin [w(T_1+T_2)]=1$

$T_1+T_2=\dfrac{\pi}{2w}\Rightarrow T_2=\dfrac{\pi}{2w}-T_1$

$\Rightarrow T_2=\dfrac{\pi}{2w}-\dfrac{\pi}{6w}=\dfrac{\pi}{3w}$

So,

$\dfrac{\pi}{3w}=2\left(\dfrac{\pi}{6w}\right)$

$\Rightarrow T_2=2T_1 \Rightarrow T_2>T_1$

Option-

$A$ is correct.

$1.00 \times 10^{20} kg$ particle is vibrating under simple harmonic motion with a period of

$1.00 \times 10^5 s$ and with a maximum speed of

$1.00 \times 10^3 m/s$ . The maximum displacement of particle from mean position is

0%
$1.59\ mm$
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$1.00\ m$
0%
$10\ m$
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$None\ of\ these$

For a SHM with given angular frequency, two arbitrary initial conditions are necessary and sufficient to determine the motion completely.These initial conditions may be

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Amplitude and initial phase
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Amplitude and total energy of oscillation
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Initial phase and total energy of oscillation
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Initial position and initial velocity

Explanation

Description of motion is completely specified if we know the variation of

$x$ as a function of time. For a simple harmonic motion, the general equation of motion is

$x=A(\omega t+\delta )$ . As

$\omega$ is given, to describe the motion completely, we need the values of

$A$ and

$\delta$ .
For option (a), we can find

$A$ and

$\delta$ if we know initial velocity and initial position. Option (c) cannot give the values of

$A$ and

$\delta$ so it is not the correct condition.

The time -period of a simple pendulum does not depend on

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length of the pendulum
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acceleration due to gravity
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mass of the bob
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none of these

Two simple harmonic motion a shown below are at right angles. They are combined to form Lissajous figures.

$x ( t ) = A \sin ( a t + \delta )$

$y ( t ) = B \sin ( b t )$ Identify the correct match below. Parameters curve

0%
$A \neq B , a = b ; \delta = 0$ Parabola
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$A = B , a = b ; \delta = \pi / 2$ line
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$A \neq B , a = b ; \delta = \pi / 2$ Ellipse
0%
$A = B , a = 2 b ; \delta = \pi / 2$ circle

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 7 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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