MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 8

Springs of spring constants K, 2K, 4K, 8K, 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillation. The time period is approximately :

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$2\pi \sqrt{\dfrac{m}{2K}}$
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$2\pi \sqrt{\dfrac{2m}{2K}}$
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$2\pi \sqrt{\dfrac{2m}{K}}$
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none of these

Explanation

In series the equation spring constant is given by:-

$\dfrac {1}{k'}=\displaystyle \sum \dfrac {1}{k'}\Rightarrow \dfrac {1}{k'}+\dfrac {1}{k}+\dfrac {1}{2k}+\dfrac {1}{4k}+\dfrac {1}{8k}+\dfrac {1}{2048k}$

$\Rightarrow \ \dfrac {1}{k'}=\dfrac {3841}{2048k}$

or approximately

$k'\simeq \dfrac {2048k}{3841}\simeq \dfrac {k}{2}$

$\therefore \$ New time period,

$T'=2\pi \sqrt {\dfrac {m}{k'}}=2\pi \sqrt {\dfrac {2m}{k}}$

A block of mass m is suspended separately by two different springs have time period

${ t }_{ 1 }$ and

${ t }_{ 2 }$ . If same mass is connected to parallel combination of both springs, then its time period is

$T$ . Then

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$T = t_1 + t_2$
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$T^2 = t_1^2 + t_2^2$
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$T^{-1} = t_1^{-1} + t_2^{-1}$
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$T^{-2} = t_1^{-2} + t_2^{-2}$

Explanation

Let the spring constant be

$k_1$ and

$k_2$

now, formula of time period of spring is given by

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

now,

$t_{1}= 2\pi \sqrt{ m/k_{1}}$

$t_{1}^{2} = \dfrac{4 \pi^{2} m}{k_{1}} \Rightarrow k_{1}=4 \pi^{2} m/t_{1}^{2}$ ----

$(2)$

$t_{2}= 2 \pi \sqrt{m/ k_{}}$

$t_{2}^{2} = \dfrac{4 \pi^{2} m}{k_{2}} \Rightarrow k_{2}= 4 \pi^{2} m/t_{2}^{2}$ ----

$(2)$

now, both are connected in parallel then ;

$k_{1}+ k_{2} \Rightarrow \dfrac{4 \pi^{2} }{t_{1}^{2}} + \dfrac{4 \pi^{2} m }{t_{2}^{2}}$

$4 \pi^{2} m \left( \dfrac{1}{t_{1}^{2}} + \dfrac{1}{t_{2}^{2}} \right)$

now, time period

$=\sqrt{\dfrac{t_{1}^{2}. t_{2}^{2}}{t_{1}^{2}+ t_{2}^{2}}}$

The bob of a

$0.2 m$ pendulum describes an arc of circle in a vertical plane. If the tension in the cord is

$\sqrt { 3 }$ times the weight of the bob when the cord makes an angle

$30$ with the vertical, the acceleration of the bob in that position is

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$9.8$
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$6.02$
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$7.2$
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$5.2$

A point mass oscillates along the

$x$ -axis according to the law

$x = x _ { 0 } \cos ( \omega t - \pi / 4 )$ . If the acceleration of the particle is written as

$a = A \cos ( \omega t + \delta )$ then:

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$A = x _ { 0 } , \delta = - \pi / 4$
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$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 4$
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$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 8$
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$A = x _ { 0 } \omega ^ { 2 } , \delta = 3 \pi / 4$

Explanation

Given,

a particle oscillation about the law,

$x=x_0\cos \left(wt-\dfrac{\pi}{4}\right)$

and acceleration :

$a=A\cos (wt+\delta)$

Since, maximum amplitude for SHM is given by :-

$a_{max}=w^2.xmax$

$(w:$ Angular frequency )

$\Rightarrow A=x_0w^2$

and, acceleration & displacement for an SHM are at opposite direction at any time

$t$ . So it results to a phase difference

$(\Delta \phi)$ of

$\pi$ between them.

$\Rightarrow \Delta \phi =(wt+\delta)-(wt-\pi/4)=\pi$

$\Rightarrow \delta =\dfrac{3\pi}{4}$ .

so, eqn for acceleration :-

$a=x_0w^2\cos \left(wt+\dfrac{3\pi}{4}\right)$ ;

$A=x_0w^2, \delta =\dfrac{3\pi}{4}$ .

A simple pendulum has time period

$T _ { 1 }.$ The point of suspension is now moved upward according to equation

$y = K t ^ { 2 }$ where

$K = 2.5 \mathrm { m } / \mathrm { s } ^ { 2 }.$ If new time period is

$T _ { 2 }$ then ratio will be

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$2 / 3$
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$5 / 6$
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$6 / 5$
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$3 / 2$

Explanation

Given,

Displacement,

$y=k{{t}^{2}}$

$\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=2k$

Since, k=1m/s²

$\dfrac{{{d}^{2}}y}{d{{t}^{2}}}={{a}_{y}}=2\,m/{{s}^{2}}$

${{T}_{1}}=2\pi \sqrt{\dfrac{l}{g}}.........(1)$

$and$

${{T}_{2}}=2\pi \sqrt{\dfrac{l}{g+{{a}_{y}}}}.........(2)$

Taking ratios of equation (1) and (2)

$\dfrac{{{T}^{2}_{1}}}{{{T}^{2}_{2}}}=\dfrac{g+{{a}_{y}}}{g}=\dfrac{10+2}{10}=\dfrac{6}{5}$

A particle undergoes SHM. When its displacement is 8 cm, it has speed 3 m/s and a speed 4 m/s at displacement of 6cm. Find time period of oscillation.

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$4\pi$
0%
$6\pi$
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$7\pi$
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$8\pi$

Explanation

using the relation

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

from given ,

$(i)$ when

$x=8,v=3$

$3=w\sqrt{A^2-64}$

$\rightarrow (1)$

$(ii)$ when

$x=6,v=4$

$4=w\sqrt{A^2-36}$

$\rightarrow (2)$

dividing

$(1)\div(2)$ given

$\left(\dfrac{3}{4}\right)^2\dfrac{A^2-64}{A^2-36}\Rightarrow A^2=\dfrac{16\times 64-9\times 36}{7}=100$

$\Rightarrow A=10\ cm$

Putting

$A$ in

$(1)$ gives,

$3=w\sqrt{10^2-b4}\Rightarrow w=\dfrac{3}{6}=\dfrac{1}{2}rad/s$

$\therefore$ Time period of oscillation,

$T=\dfrac{2\pi}{w}=4\pi$

The time period of oscillation of a

$SHM$ is

$\frac { \pi } { 2 } s.$ Its acceleration at a phase angle

$\frac { \pi } { 3 }\ rad$ from extreme position is

$2 m s ^ { - 2 }.$ What is its velocity at a displacement equal to half of its amplitude from mean position? (in

$m s ^ { - 1 }$ )

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$0.707$
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$0.866$
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$\sqrt { 2 }$
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$\sqrt { 3 }$

If function

$\sin^{2}(\omega t)$ represents position of a particle from the origin as function of time, then motion of particle is

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A simple harmonic motion with a period $2\pi/\omega$
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A simple harmonic motion with a period $\pi/\omega$
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A period motion but not simple harmonic motion with a period $2\pi/\omega$
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A period motion but not simple harmonic motion with a period $\pi/\omega$

If the maximum velocity of a particle in SHM is

${ v }_{ 0 }$ . then its velocity at half the amplitude from position of rest will be :

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${ v }_{ 0 }/2$
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${ v }_{ 0 }$
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${ v }_{ 0 }\sqrt { \dfrac{3}{2} }$
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$\dfrac{{ v }_{ 0 }\sqrt { 3 }}{2}$

A block of mass

$10\,Kg$ is suspended through two light spring balances as shown in figure.

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Both the scales will read $10\,Kg$
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Both the scales will read $5\,Kg$
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The upper scale will read $10\,Kg$ and the lower zero
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The readings may be anything but their sum will be $10\,Kg$

Explanation

Since the block and the balances are in equilibrium and considering that balances are massless.

$\Rightarrow T_1 = 10kg$ and

$T_2 = T_1$

$\Rightarrow T_ = T_2 = 10kg$

and, hence, both the balances will read 10kg.

1524125_1205988_ans_8a413c5f4e52480fa4d9e54eac7358d2.png

A particle is executing S.H.M. with amplitude

$'a'$ and has maximum velocity

$'v'$ . Its speed at displacement

$a/2$ will be

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$0.866\ v$
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$v/2$
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$v$
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$v/4$

A particle moves along a straight line such that its displacement(s) at any time(t) is given by

$S=t^3-3t^2+2$ . The velocity of the particle when its acceleration zero is?

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$-2$ units
0%
$-8$ units
0%
$0$ units
0%
$-9$ units

Explanation

Displacement is given as $=({ t }^{ 3 }−3{ t }^{ 2 }+2)\Longrightarrow (1)$

We know that $a=\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } }$

Differentiating (1 ) once with respect to t we get

$\dfrac { ds }{ dt } =\dfrac { d }{ dt } ({ t }^{ 3 }−3{ t }^{ 2 }+2)=3{ t }^{ 2 }−6t$

Differentiating once again we get

$\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } } =\dfrac { d }{ dt } (3{ t }^{ 2 }−6t)=6t−6$

Given condition is $\dfrac { { d }^{ 2 }s }{ d{ t }^{ 2 } } =6t−6=0$

$t=1s$

$s(1)=(13−3×12+2)=0m$

The time period of a particle in simple harmonic motion is equal to the time between consecutive apperance of the particle at a particular point in its motion. This point is

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The mean position
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An extreme position
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Between the mean position and the positive extreme
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Between the mean position and the negative extreme

A particle executing

$SHM$ while moving from the one extremity is found at distance

$x_1, x_2$ and

$x_3$ from the centre at the end of three successive seconds. The time period of oscillation is

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

The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in figure. The time period of oscillation of particle is

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

Explanation

Row Hg

$V_{max} = Aw = 0.um/s$

$x_{max} = A = 10$

$\frac{Aw}{A} = \frac{0.u}{10\times 10^{-2}} = \frac{40}{10} = 4$

$w = 4$

$T = \frac{2\pi }{w} = \frac{2\times \pi }{4}$

$\therefore T = \pi /2^s$

Answer is A

1191439_1191262_ans_64fe3b1ba40a409cb0fa1df8cb1d37c6.JPG

A simple pendulum has the time period of T at the surface of Earth. if it is taken to a height equal to R above the surface of Earth, then the new time period will be :

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

A cubical body of side

${10}^{-1}m$ and mass

$2\times {10}^{-3}kg$ is floating in water. When it is pressed inside water and then released. It executes SHM its time period will be

0%
$28\ sec$
0%
$0.028\ sec$
0%
$0.28\ sec$
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$2.8\ sec$

A stone falls freely from rest a height

$h$ and it travels a distance of

$\dfrac{h}{2}$ in the last second.The time of journey is

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$\sqrt 2s$
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$(2- \sqrt 2)s$
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$(2+\sqrt 2)s$
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$2s$

The differential equation representing the SHM of a particuleis

$\cfrac{9d^2y} {dt^2}+4y=0$ . The time period of the particle is given by

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$\cfrac\pi 3 {sec}$
0%
$\pi {sec}$
0%
$\cfrac2\pi 3 {sec}$
0%
$3\pi {sec}$

Explanation

Given,

$9\dfrac{d^2y}{dt^2}+4y=0$ . . . .. . . . . .(1)

acceleration,

$a=\dfrac{d^2y}{dt^2}$

equation (1) becomes,

$9a+4y=0$

$a=-\dfrac{4}{9}y$ (

$a=-\omega ^2y$ )

$\omega ^2=\dfrac{4}{9}$

$\omega =\dfrac{2}{3}$

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

$T=3\pi sec$

The correct option is D.

The force of a required to row a boat at velocity is proportional to square of its speed of v km/ h requires 4 KW, how many does a seepd of 2V km/h required

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8 kW
0%
16 kW
0%
32kw
0%
76kw

Explanation

Here

$F\propto V^{2}$

$P = F\times V$

$\propto V^{2}\times V = V^{3}$

$\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{V_{1}^{3}}{V_{2}^{3}}=\left(\dfrac{V}{2V}\right)^{3}=\dfrac{1}{8}$

$\therefore P_{1}=4kw$

$P_{2}=8\times 4= 32kw$

$\therefore P_{2}= 32kw$

The differential equation representing the SHM of a particule is

$\dfrac { { 9d }^{ 2 }y }{ { dt }^{ 2 } }$

$+ 4y = 0.$ The time period of the particle is given by :

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

Explanation

The differential equation representing S.H.M of a particle is,

$9\dfrac{d^2y}{dt^2}+4y=0$

$9a+4y=0$

$a=-\dfrac{4}{9}y$

$\omega ^2=\dfrac{4}{9}$ (

$a=-\omega ^2 y$ )

$\omega =\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}$

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

$T=3\pi sec$

The correct option is D.

Spring in vehicles are introduces to:

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Reduce
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Reduce impluse
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Reduce force
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Reduce velocity

A particle executing

$S.H.M$ of amplitude

$4\ cm$ and

$T=4\ sec$ . The time taken by it move from positive extreme position to half the amplitude is:-

0%
$1\ sec$
0%
$\dfrac {1}{3}\ sec$
0%
$\dfrac {2}{3}\ sec$
0%
$\sqrt{\dfrac {2}{3}}\ sec$

Explanation

$y$ is the displacement,

$t$ the time,

$a$ the amplitude and

$\omega$ the angular frequency

then

$y=a\cos{\omega t}$

Given

$y=\dfrac{a}{2}\Rightarrow \dfrac{a}{2}=a\cos{\omega t}$

$\Rightarrow \cos{\omega t}=\dfrac{1}{2}$

$\Rightarrow \dfrac{1}{2}=\cos{\dfrac{2\pi}{T}\times 2}$ at

$T=4$ secs

$\Rightarrow t\cos{\dfrac{2\pi}{4}}=\dfrac{1}{2}$

Let

${W}_{A}$ be the work done by the spring

$A$ and

${W}_{B}$ be the work done by the spring

$B$

$\Rightarrow \dfrac{{W}_{A}}{{W}_{B}}=\dfrac{{K}_{A}}{{K}_{B}}=\dfrac{1}{3}$

$\therefore {W}_{A}:{W}_{B}=1:3$

A particle oscillating in simple harmonic motion is :

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Never in equilibrium because it is in motion
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Never in equilibrium because there is always a force
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In equilibrium at the ends of its path because of zero velocity there
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In equilibrium at the centre of its path because the acceleration is zero there

A particle is executing

$S.H.M.$ If the amplitude is

$2m$ and periodic time

$2 \ \ seconds$ , then the max. velocity of the particle will be

0%
$\pi m/s$
0%
$\sqrt 2 \pi m/s$
0%
$2 \pi m/s$
0%
$4 \pi m/s$

Explanation

In an

$S.H.M$

$A=2\ m, T=2\ s$

So,

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

$V_{max}=A\omega$

$=2\pi\ m/s$

A spring-block system undergoes simple harmonic motion on a smooth horizontal surface. The block is now given some positive charge, and a uniform horizontal electric field to the right is switched on. As a result,

0%
the time period of oscillation will increase
0%
the time period of oscillation will decrease
0%
the time period of oscillation will remain unaffected
0%
the mean position of simple harmonic motion will shift to the left

A particle performs SHM with period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, starting from right extreme position is

0%
$0$
0%
$\dfrac{2A}{T}$
0%
$\dfrac{4A}{T}$
0%
$\dfrac{3A}{T}$

Explanation

Let the displacement of the particle executing SHM at any instant is

$x=Asinwt$

Velocity

$v=\dfrac { dx }{ dt } =\dfrac { d }{ dt } \left( Asinwt \right) =Awcoswt$

The mean velocity of the particle average over quarter oscillation is

${ \left< v \right> }_{ 0\rightarrow T/4 }=\dfrac { \int _{ 0 }^{ T/4 }{ v } dt }{ \int _{ 0 }^{ T/4 }{ dt } } =\dfrac { \int _{ 0 }^{ T/4 }{ Aw\cos } dt }{ T/4 }$

$=\dfrac { \dfrac { Aw }{ w } { \left[ sinwt \right] }_{ 0 }^{ T/4 } }{ T/4 }$

$=\dfrac { A{ \left[ sinwt \right] }_{ 0 }^{ T/4 } }{ T/4 }$

$=\dfrac { 4A }{ T }$

A body of mass

$100 gm$ is performing

$S.H.M$ with a period of

$4.5s$ and amplitude

$7 cm$ . Velocity of the body at equilibrium position in

$cm/s$ is

0%
$10$
0%
$8$
0%
$6$
0%
$5$

A particle executes SHM of type

$x=asin\omega t$ . It takes time

$t_1$ from

$x=0$ to

$x=\dfrac{a}{2}$ and

$t_2$ from

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

$t_2$ from

$x=\dfrac{a}{2}$ to

$x=a$ . The ratio of

$t_1:t_2$ will be:

0%
$1:1$
0%
$1:2$
0%
$1:3$
0%
$2:1$

Explanation

By symmetry the time period from

$a=0$ to

$a=a$ is

$T/4$ . Let eliminate the options the time taken at extreme i.e. at

$n=a$ the time taken is more at extreme than at mid position so

${ t }_{ 2 }>{ t }_{ 1 }$

$\dfrac { a }{ 2 } =asinw{ t }_{ 1 }\Rightarrow { wt }_{ 1 }=\dfrac { \pi }{ 6 } \Rightarrow$

${ t }_{ 1 }=\dfrac { \pi }{ 6 } \times \dfrac { T }{ 2\pi } =\dfrac { T }{ 12 }$

${ t }_{ 2 }=\dfrac { T }{ 4 } -\dfrac { T }{ 12 } =\dfrac { T }{ 6 }$

$\dfrac { { t }_{ 2 } }{ { t }_{ 1 } } =\dfrac { 1 }{ 2 }$

An object exceuting SHM has an acceleration of

$0.2m/s^{2}$ at a distance of

$0.05$ m from the mean position calculate the period of SHM

0%
$2.1435$
0%
$3.1428$
0%
$1.432$
0%
$4.321$

Explanation

Given,

$a=0.2m/s^2$

$x=0.05m$

The time period of S.H.M is given by

$a=\omega ^2 x$

$a=(\dfrac{2\pi}{T})^2 x$ (

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

$T^2=4\pi^2 \dfrac{x}{a}$

$T=2\pi \sqrt{\dfrac{x}{a}}$

$T=2\times 3.1428\times \sqrt{\dfrac{0.05}{0.2}}$

$T=3.1428 sec$

The correct option is B.

The acceleration of a particle in

$SHM$ at

$5\ cm$ from its mean position is 20

$cm/{ sec }^{ 2 }$ . The value of angular frequency in radians/ sec will be :

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

The amplitude and time period of a particle performing

$S.H.M$ are ;a;and

$T$ respectively. The displacement at which its velocity will be half the max. velocity will be

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

Explanation

The displacement of a point

$\times$ subjected to sumis

$x=asinT$

The velocity

$V$ of that point

$V=acosT$

When

$V=\dfrac { acosT }{ 2 }$

$\Rightarrow cos\left( 0.5 \right) =600$

At this velocity

$\left( V={ 60 }^{ 0 } \right)$ Displacement

$x=a\sin{ 60 }^{ 0 }=0.866a$

The position vector of a particle from origin is given by

$\hat r = A\left( {\hat i\cos \omega t + \hat jsin\omega t} \right)$ . The motion of the particle is

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simple harmonic
0%
on a straight line
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on a circle
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with contact acceleration

A particle is subjected to two simple harmonic motions given by:

$y_1=10 \sin\omega t$ and

$y_2=5 \sin (\omega t+ \pi)$ The maximum speed of the particle is:

0%
$\sqrt{10^2+5^2 \omega}$
0%
$\sqrt{10^2-5^2 \omega}$
0%
$5\omega$
0%
$15\omega$

Explanation

$y = {y_1} + {y_2}$

$\Rightarrow y = 10\sin wt + 5\left( {\sin \omega t + \pi } \right)$

$\Rightarrow y = 5\sin \omega t$

$\therefore {V_{\max }} = 5 \omega$

hence,

option

$(C)$ is correct answer.

1201176_1235065_ans_eb9018dcaebe481f91f498c63f4630d8.PNG

The maximum time period of oscillation of a simple pendulum of larger length is :

0%
$Infinity$
0%
$24\ hours$
0%
$12\ hours$
0%
$1\ 1/2\ hours$

$\upsilon$ as shown in figure. The velocity of the block on the plane 'u' is

0%
$\upsilon$ cos $ec \theta$
0%
$\upsilon$ sin $\theta$
0%
$\upsilon$ cos $\theta$
0%
$\upsilon$ sec $\theta$

A particle of mass

$2$ kg, executing SHM has amplitude

$20$ cm and time period

$1$ s. Its maximum speed is

0%
$0.314$ m/s
0%
$0.628$ m/s
0%
$1.256$ m/s
0%
$2.512$ m/s

Explanation

$v_{max}=Aw$

$=20cm \times \dfrac{2\pi}{T}$

$=0.2m \times \dfrac{2\pi}{1s}$

$=1.256m/s$

A particle executes SHM of time period T. It takes time

$t_1$ to go from

$x=0$ to

$x = \frac{a}{2}$ and

${t_2}$ from

$x = \frac{a}{2}$ to

$x=0$ . The ratio of

${t_1}:{t_2}$ will be:

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

Two particles executes S.H.M along the same line at the same frequency. They move in opposite direction at the mean position. The phase difference will be :

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

Explanation

During SHM the position of particle is given by

$x = A\sin wt$

Let the position 1st particle is given by

$x = A_1\sin (wt)$

and that of 2nd [article

$x = A_2\sin (wt + \phi)$

as frequency of both partical is same

the position of mean

$\Rightarrow x=0$

for 1st particle (A)

$0 = A\sin wt$

$wt = 0$ ...(1)

$t=0$ at mean position

for 2nd particle (B)

$0 = \sin (wt + \phi)$

$wt + \phi = 0$ or

$\pi$

as if move in opposite direction

$wt + \phi = \pi$

as form (1)

$wt = 0$

$\Rightarrow \phi = \pi$

Hence phase difference

$= \pi$

1463815_1236103_ans_2436f2995fa843159cb4d0c465872fa4.png

A particle has simple harmonic motion. The equa-tion of its motion is

$x=5\sin (4t- \pi / 6)$ , where

$x$ is its displacement. If the displacement of the particle is

$3\ units$ , then its velocity is

0%
$2 \pi /3$ units
0%
$5 \pi /6$ units
0%
$20$ units
0%
$16$ units

Explanation

From the given equation,

The amplitude of the wave is

$a=5$

and the angular velocity of the wave is

$\omega =4$

The velocity of the wave can be obtained as:

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

$=4\sqrt{(5)^2-(3)^2}=16\ units$

The maximum speed of a particle executing

$SHM$ is

$1\ m/s$ and maximum acceleration is

$1.57m/s^{2}$ . Its time period is

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

Explanation

Maximum speed

$V = Aw$

So,

$1 = Aw$

Also, maximum acceleration

$a = Aw^2$

So,

$1.57 = Aw^2$

$\dfrac{1.57}{1} = \dfrac{Aw^2}{Aw}$

$\implies \ w= 1.57$

So,

$\dfrac{2\pi}{T} = 1.57$

$\implies \ T = \dfrac{2\pi}{1.57} = 4 \ s$

If a hallow spherical vessel is slowly filed up to middle with water, then its period of oscillation

0%
first increases then decreases
0%
Remains same
0%
increases
0%
decreases

A particle starts simple harmonic motion from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3E/Its displacement at the instat is-

0%
$a/\sqrt { 2 }$
0%
$a/2$
0%
$\dfrac { a }{ \sqrt { 3/2 } }$
0%
$a/\sqrt { 3 }$

Explanation

The equation of the simple harmonic motion is

$y=A sin \omega t$ with usual notations.

The total energy is

$\left( {\frac{1}{2}} \right)m{\omega ^2}{a^2}$ and the kinetic energy is

$\left( { \frac { 1 }{ 2 } } \right) m{ \omega ^{ 2 } }\left( { { a^{ 2 } }-{ y^{ 2 } } } \right) .$

Therefore we have

$\left( {\frac{1}{2}} \right)m{\omega ^2}{a^2} = E$ and

$\left( {\frac{1}{2}} \right)m{\omega ^2}\left( {{a^2} - {y^2}} \right) = \dfrac{{3E}}{4}.$

Dividing the second equation by the first ,

$1 - \dfrac{{{y^2}}}{{{a^2}}} = \dfrac{3}{4}$ from which

$y=\dfrac{a}{2}$ .

A particle is executing

$SHM$ along

$x$ -axis given by

$x = A \sin \omega t$ . The average velocity during the time interval

$t = 0$ to

$t = T$ is

0%
${4A}/{T}$
0%
$0$
0%
${2A}/{T}$
0%
${A}/{T}$

A wave has SHM whose period is 4 s while another wave which also possess SHM has its period 3 s. If both are combined, then the resultant wave will have the period equal to:

0%
4 s
0%
5 s
0%
12 s
0%
3 s

Time period of a particle executing

$SHM$ is

$8\ sec$ . At

$t=0$ it is at the mean position. The ratio of the distance covered by the particle in the

$1st$ second to the

$2nd$ second is :

0%
$\dfrac{1}{\sqrt{2}+1}$
0%
$\sqrt{2}$
0%
$\dfrac{1}{\sqrt{2}}$
0%
$\sqrt{2}+1$

Explanation

${Y}_{1}=A\sin{\omega}=A\sin{\dfrac{2\pi}{8}}=\dfrac{A}{\sqrt{2}}$

Displacement when

$t=2$ s is

${Y}_{2}=A\sin{2\omega=A\sin{2\times\dfrac{\pi}{4}}}=A\sin{\dfrac{\pi}{2}}=A$

Displacement covered in time

$2$ s

$=A-\dfrac{A}{\sqrt{2}}$

Ratio of these two displacement when

$t=1$ s and

$t=2$ s

$\dfrac{{Y}_{1}}{{Y}_{2}}=\dfrac{\dfrac{A}{\sqrt{2}}}{A-\dfrac{A}{\sqrt{2}}}$

$=\dfrac{\dfrac{A}{\sqrt{2}}}{A\left(1-\dfrac{1}{\sqrt{2}}\right)}$

$=\dfrac{1}{\sqrt{2}-1}$

$=\dfrac{1}{\sqrt{2}-1}\times\dfrac{\sqrt{2}+1}{\sqrt{2}+1}$

$=\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}+1$

A string of mass m is fixed at both ends. The fundamental tone oscillations are excited in the strong with angular frequency

$\omega$ and maximum displacement amplitude A. Find the total energy curtained in the string.

0%
$\frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }$
0%
$\frac { 1 }{ 4 } { m\omega }^{ 2 }{ A }^{ 2 }$
0%
$\frac { 1 }{ 6 } { m\omega }^{ 2 }{ A }^{ 2 }$
0%
$\frac { 1 }{ 8 } { m\omega }^{ 2 }{ A }^{ 2 }$

Explanation

$y=A\sin kx\cos \omega t$ where

$k=\dfrac{2\pi}{\lambda}$

$\dfrac{dy}{dt}=-A\omega \sin kx\sin \omega t$

Kinetic energy of element dx at position x is

$dk_n=\dfrac{1}{2}\mu dx\left(\dfrac{dy}{dt}\right)^2$

$=\dfrac{1}{2}\mu A^2\omega^2\cos^2\omega t\cdot \sin^2kx.dx$

$k_n=\displaystyle\int^{x=1}_{x=0}dk_n=\dfrac{1}{4}MA^2\omega^2\cos^2\omega t$ .

1221071_1259876_ans_7df1956fb20942b39249f3f2fc4caa9d.png

A dog of mass

$m$ is walking on a pivoted disc of radius

$R$ and mass

$M$ in a circle of radius

$R/2$ with an angular frequency

$n$ . The disc will revolve in opposite direction with frequency :

0%
$\dfrac{mn}{M}$
0%
$\dfrac{mn}{2M}$
0%
$\dfrac{2 mn}{M}$
0%
$\dfrac{2Mn}{M}$

Explanation

Angular momentum of the system should be zero.

Moment of inertia of the dog is

$m{\left( {\dfrac{R}{2}} \right)^2}$

Angular velocity

$=n$

$\therefore$ Angular momentum

$L = IW = m\dfrac{{{R^2}}}{4}n.......\left( 1 \right)$

Moment of inertia of disc

$= \frac{1}{2}M{R^2}$

Let the disc be rotating with angular velocity

$w$

$\therefore$ Angular momentum

$= \frac{1}{2}M{R^2}w$

from

$(1)$ and

$(2)$ we get,

$\frac{1}{2}M{R^2}w = \frac{1}{4}m{R^2}n$

$w = \dfrac{{mn}}{{2M}}$

A particle executes simple harmonic motion between x = - A and x = + A .The time taken by it to go from O to A /2 is T

$_{1}$ and to go from A /2 to A is T

$_{2}$ . Then

0%
T $_{1}$ < T $_{2}$
0%
T $_{1}$ > T $_{2}$
0%
T $_{1}$ = T $_{2}$
0%
T $_{1}$ = 2 T $_{2}$

Explanation

(a) The equation of SHM, when it starts swinging from the mean position, is

$y=Asin\omega{t}$ .

At mean position, y=0, t=0 .....(1)

$\dfrac{A}{2}=Asin\omega{t_2}$ or

$sin\omega{t_2}=\dfrac{1}{2}=sin\dfrac{\pi}{6}$

$\omega{t_2}=\dfrac{\pi}{6}$ or

$t_2=\dfrac{\pi}{6\omega}$ .....(2)

At y=A time for travel from 0 to A=

$t_3$

A=Asin

$\omega{t_3}$

or sin

$\omega{t_3}=1=sin\dfrac{\pi}{2}$ or

$t_3=\dfrac{\pi}{2\omega}$ .....(3)

$T_1= t_2-t_1=\dfrac{\pi}{6\omega}$ .....(4)

$T_2=t_3-t_2=\dfrac{\pi}{2\omega}-\dfrac{\pi}{6\omega}=\dfrac{\pi}{3\omega}$ .....(5)

$\dfrac{T_1}{T_2}={\dfrac{\dfrac{\pi}{6\omega}}{\dfrac{\pi}{3\omega}}}=\dfrac{3\omega}{6\omega}=\dfrac{1}{2}$ or

$\dfrac{T_1}{T_2}=\dfrac{1}{2}$

$T_1<T_2$

A uniform straight rod of mass 'm; kg and length L is hinged at one end. It is free to oscillate in vertical plane. A point of mass (m kg) is attached to it at a distance 'x' from the hinge. The value of x for which time period of oscillations will be minimum is (use:

$\sqrt { \frac { 7 }{ 3 } =1.5 }$ )

0%
$2L/3$
0%
$L/2$
0%
$L/4$
0%
$2L/7$

Explanation

For a rigid body oscillating about horizontal axis passing through the point O the time period is given by

$T=2\pi \sqrt{\dfrac{l}{mg(l in)}}$

$=$ length of centre of gravity.

For real centre

$=l/2$ from ringed point

For mass of x

Centre of mass of system

$c=\dfrac{mx+ml/2}{2m}$

$=\dfrac{(2l+l/2)}{2}$

Now,

Moment of inertia of system

$=\dfrac{1}{3}ml^2+mx^2$

$T=2\pi \sqrt{\dfrac{\dfrac{1}{3}ml^2+mx^2}{(2mg)\left(\dfrac{x+l/2}{2}\right)}}$

T should be minimum then

$\dfrac{\dfrac{1}{3}ml^2+mx^2}{(2mg)\left(\dfrac{x+l/2}{2}\right)}$ should be minimum

Let

$=\dfrac{\dfrac{1}{3}l^2+x^2}{\left(\dfrac{x}{3}+\dfrac{l}{4}\right)}$

Should be minimum

$x=\dfrac{L}{4}$ .

1223011_1264807_ans_8a85ea81a85b43ea94593a47d8b046ec.png

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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