MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 9

In the arrangement as shown, the length of rod is L and mass is M and it is connected to smooth pin at O.The rod is placed on smooth horizontal table.The spring constant of both the spring is K.The rod is displaced by a small angle

$\theta$ to perform small ocillation, then is given by

0%
$\left( \cfrac { 3K }{ M } \right) ^{ 1/2 }$
0%
$\left( \cfrac { 6K }{ M } \right) ^{ 1/2 }$
0%
$\left( \cfrac { 4K }{ M } \right) ^{ 1/2 }$
0%
$\left( \cfrac { 3K }{ 2M } \right) ^{ 1/2 }$

A simple harmonic wave in a medium is given by,(y is in centimeters)

$y=\frac { 10 }{ \pi } sin (2000{ \pi }t- \frac { \pi x }{ 17 } )$
The maximum velcocity of a particle executing tha wave is

0%
$330ms^-1$
0%
$20ms^-1$
0%
$200ms^-1$
0%
$2000ms^-1$

A mass is suspended from a wire and is pulled along the length of wire, resulting in oscillations of time period

$T_{1}$ . The same mass is next attached to wire of the same material and length but double the cross-sectional area. The time period this time

$T_{2}$ . Then

$T_{1}/T_{2}$ is equal to

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

Oscillatory or vibrating motion means to and fro motion

0%
Which does not repeat
0%
which repeats after unequal interval of time
0%
which repeats after equal interval of time
0%
which remains unchanged

Two particles

$P$ and

$Q$ start from origin and execute simple harmonic motion along

$X-$ axis with same amplitude but with periods

$3\ s$ and

$6\ s$ respectively. The ratio of the velocities of

$P$ and

$Q$ when they meet is

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

Explanation

The particles will meet at the mean position when

$P$ completes one oscillation and

$Q$ completes half an oscillation

The angular velocity for first particle is

$\omega_1=\dfrac{2\pi}{3}$ .

The angular velocity for second particle is

$\omega_2=\dfrac{2\pi}{6}=\dfrac{\pi}{3}$

So, the velocities at different points will be:

$v_1=\omega_1A$ and

$v_2=\omega_2A$

So,

$\dfrac{v_1}{v_2}=\dfrac{\omega_1}{\omega_2}$

$\dfrac{v_1}{v_2}=\dfrac{2\pi/3}{\pi/3}$

$\dfrac{v_1}{v_2}=2:1$

Time period of small oscillation (in a vertical plane normal to the plane of strings) of the bob in the arrangement shown will be

0%
$2\pi \sqrt{ \cfrac{l}{g}}$
0%
$2\pi \sqrt {\cfrac {l}{\sqrt {2g}}}$
0%
$2\pi \sqrt {\cfrac {\sqrt {2l}}{g}}$
0%
$2\pi \sqrt {\cfrac {2l}{g}}$

Explanation

$\textbf{Hint:}$ Use the time period formula

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

$\textbf{Step 1: Calculating effective length}$

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

Given values are

$L = ab$

$\cos 45^{\circ} = \dfrac{ab}{l} = \dfrac{1}{\sqrt{2}}$

$ab = \dfrac{l}{\sqrt{2}}$

$\textbf{Step 2: Calculating time period}$

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

$\textbf{Option B is correct.}$

A particle is executing SHM about

$\gamma =0$ along y- axis.Its position at an instant is given by

$\gamma =(7m)$ sin(πt) its average velcocity for a time interval 0 to 0.5 s is

0%
$14m/s$
0%
$7m/s$
0%
$\dfrac { 1}{ 7 } m/s$
0%
$28m/s$

Explanation

at t=0, position of particle , r=0

at t=0.5, position of particle, r=7m

Total displacement = 7m

Total time = 0.55

$Average\,velocity =\frac{Total\, displacement}{Total\, time}$

$= \frac{7}{0.5}= 14\, m/s$

1222097_1279864_ans_4df1582c52f54b4c9ab68ea9564a768b.jpg

The velocity v of a particle of mass m moving along a straight line changes with time 't' as

$\dfrac{{{d^2}v}}{{d{t^2}}} = Kv\;$ where 'K' is a positive constant. Which of the following statement is correct:

0%
The particle does not perform SHM
0%

The particle performs SHM with time period $2\pi \sqrt {\dfrac{m}{k}}$
0%
nothing occurs
0%
its in non moving state

The time period of a particle executing

$SHM$ is

$8\ s$ . At

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

${ 1 }^{ st }$ second to the

${ 2 }^{ nd }$ second is

0%
$(\dfrac{1}{\sqrt { 2 } -1})$
0%
$\sqrt { 2 }$
0%
$(\sqrt { 2 } +1)$
0%
$\dfrac { 1 }{ \sqrt { 2 } }$

Explanation

$\begin{array}{l} Y'=A\sin w=A\sin \dfrac { { 2\pi } }{ 8 } =\dfrac { A }{ { \sqrt { 2 } } } \\ Dis\tan ce\, \, when\, \, t=2\, \, \sec \, \, is \\ { Y^{ 2 } }=A\sin w=A\sin \frac { \pi }{ 2 } =A \\ Dis\tan ce\, \, { { cov } }ered\, \, in\, \, 2\sec \, \, is\, \, \dfrac { { A-A } }{ { \sqrt { 2 } } } \\ \dfrac { { { Y^{ 1 } } } }{ { { Y^{ 2 } } } } =\dfrac { { \left( { \dfrac { A }{ { \sqrt { 2 } } } } \right) } }{ { \left( { A-\dfrac { A }{ { \sqrt { 2 } } } } \right) } } \\ =\dfrac { 1 }{ { \sqrt { 2 } -1 } } \end{array}$

Initially the spring is undeformed. Now the force

$F$ is applied to

$B$ as shown. When the displacement of

$B$ w.r.t

$A$ is

$x$ towards right in some time then the relative acceleration of

$B$ w.r.t

$A$ at the moment is:

0%
$\cfrac{F}{2m}$
0%
$\cfrac{F-kx}{m}$
0%
$\cfrac{F-2kx}{m}$
0%
$none\ of\ these$

Explanation

A moment is

$\frac{{F - 2kx}}{m}$ .

Option

$C$ is correct answer.

Two particles of same period

$(T)$ and amplitude undergo

$SHM$ along the same line with initial phase of

$\pi/6$ . If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:

0%
$3T/8$
0%
$T/4$
0%
$T/2$
0%
$T$

Explanation

$x=A\sin \left ( wt-\dfrac{\pi}{2} \right )$

$\sin \left ( wt-\dfrac{\pi}{2} \right )=\sin wt$

$2wt-\dfrac{\pi}{2}=\pi$

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

$wt=\dfrac{3}{4}\pi$

$t=\dfrac{3}{4}\dfrac{\pi}{w}$

$t=\dfrac{2\pi}{w}\left ( \dfrac{3}{8} \right )$

$\therefore t=\dfrac{3T}{8}$

The total energy of a particle executing SHM is directly proportional to the square of the following quantity.

0%
Acceleration
0%
Amplitude
0%
Time period
0%
Mass

A particle executes

$SHM$ and its position with time as

$x = A \sin\omega t$ . Its average speed during its motion from mean position to mid-point of mean and extreme position is

0%
$Zero$
0%
$\cfrac{3A\omega}{\pi}$
0%
$\cfrac{A\omega}{2\pi}$
0%
$\cfrac{2A\omega}{\pi}$

A particle executes SHM according to equation $x = 10 \cos[2\pi t+\pi / 2]$ , where $t$ is in second. The magnitude of the velocity of the particle at $t=1/6s$ will be:

0%
$2.47$
0%
$2.86$
0%
$2.05$
0%
$31.4$

Two charges each $+5\mu C$ are at $(0, \pm 3)$ . A third charge of $1\mu C$ and of mass $2g$ is at $(4, 0)$ . What is the minimum velocity to be given to the $1\mu C$ charge such that it just reaches the origin (in $ms^{-1}$ )?

0%
$3$
0%
$12$
0%
$\sqrt12$
0%
$4$

A particle undergoes

$SHM$ of period

$T$ . The time taken to complete

$3/8th$ oscillation starting from the mean position is

0%
$\dfrac{3}{8}T$
0%
$\dfrac{5}{8}T$
0%
$\dfrac{5}{12}T$
0%
$\dfrac{7}{12}T$

Explanation

Time to complete

$\dfrac{1}{8}$ th oscillation from extreme position is,

$y=\dfrac{a}{2}=a\cos wt=a \cos \dfrac{2\pi}{T}t$

$\Rightarrow \cos \dfrac{\pi}{3}=\cos \dfrac{2\pi t}{T}$

$\Rightarrow t=\dfrac{T}{6}$

w.k.t time taken to complete one fourth oscillation is

$\dfrac{T}{4}$

Therefore the time to complete

$\dfrac{3}{8}$ th oscillation

$=\dfrac{T}{4}+\dfrac{T}{6}$

$=\dfrac{5T}{12}$

The mass of particle executing S.H.M is 1 gm.If its periodic time is

$\pi$ seconds, the value of force constant is:-

0%
4 dynes/cm
0%
4 N/cm
0%
4 N/m
0%
4 dynes/m

Explanation

Given, mass

$m=1 gm$ ; time period

$T=\pi$ sec

The angular frequency

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

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

$k$ is the force constant),

$k=m\omega^2=(1 gm)(2)^2=4$ dynes/cm

A particle is executing SHM about y=0 along y-axis. Its position at an instant is given by y=(7m) sin

$\left( \pi f \right) .$ its average velocity for a time interval 0 to 0.5 s is

0%
14 m/s
0%
7 m/s
0%
$\dfrac { 1 }{ 7 } m/s$
0%
28 m/s

A small body of mass

$0.10 kg$ is executing S.H.M of amplitude

$1.0m$ and period

$0.20sec$ . The maximum force acting on it is:

0%
$98.596N$
0%
$985.96N$
0%
$100.2N$
0%
$76.23N$

Explanation

Given:

The mass of the body is

$m=0.10\ kg$

The amplitude of SHM

$a=1.0\ m$ .

The time period of oscillation is

$T=0.20\ s$

Maximum acceleration of the syustem is given by:

$A_{max}=a \omega^2$

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

$=\dfrac{a \times 4 \pi^2}{T^2}=\dfrac{1 \times 4 \times (3.14)^2}{0.2 \times 0.2}$

So, the maximum force acting on the body will be:

$F_{max}= m \times A_{max}=\dfrac{0.1 \times 4 \times (3.14)^2}{0.2 \times 0.2}=98.596N$

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

$5$ cm along x axis, origin as mean position. If at

$x= +4$ cm magnitude of velocity is equal to magnitude of acceleration then find time period of oscillation in seconds.

0%
$\frac{4\pi}{3}$
0%
$6\pi$
0%
$\frac{8\pi}{3}$
0%
$\frac{9\pi}{2}$

A simple harmonic oscillator of angular frequency $2$ rad/s is acted upon by an external force $F = \sin t$ N. If the oscillator is at rest in its equilibrium position at $t= 0$ , its position at later times is proportional to:

0%
$\sin t + \cfrac{1}{2} \cos 2t$
0%
$\cos t - \cfrac{1}{2} \sin 2t$
0%
$\sin t + \cfrac{1}{2} \sin 2t$
0%
$\sin t - \cfrac{1}{2} \sin 2t$

Explanation

According to the question,

Given;

$F= Fsin t \Rightarrow a = \dfrac{F}{m} sin t \Rightarrow v = \dfrac{-F}{m} cos t$

$x = \dfrac{-F}{m} sin t \Rightarrow X = r \vec{x_1} + r \vec{x_2} = Asin 2t - \dfrac{F}{m} sint$ ..............................

$eq-i$

Now, we know that in

$SHM$ ,

$a \ ( acceleration) = - \omega^2 A sin \omega t$ and above we find

$a = \dfrac{F}{m} sin t$

Resultant force,

$F_{resultant} = sint - m \omega^2 A sinwt$

$\Rightarrow F_R = sint - 4mAsin2t$ (Given,

$\omega = 2 radian/s$ )

$\Rightarrow a_R = sint /m- 4Asin2t$ (By integrating)

$\Rightarrow v_R = -cost /m + 2Acos2t$ (By integrating)

$\Rightarrow x_R = -sint /m + Asin2t$

We can write,

$x_R = \dfrac{-1}{m} (sint - Amsin2t) \ \Rightarrow x_R \propto (sint - Amsin2t)$

Only option

$D$ is in the given form, Hence, option

$D$ is correct.

A student says that he had applied a force

$F= -k\sqrt {x}$ on a particle and the particle moves in simple harmonic motion. He refuses to tell whether

$k$ is a constant or not. Assume that he has worked only with positive

$x$ and no other force acted on the particle.

0%
As $x$ increases $k$ increases
0%
As $x$ increases $k$ decreases
0%
As $x$ increases $k$ remains constant
0%
The motion cannot be simple harmonic

Explanation

Correct Answer: Option A
Step 1: Use force law for SHM.
As the motion of the particle is Simple Harmonic Motion.

From force law,

$F = -kx$

where

$k$ is a constant

$= m\omega^2$

$F=-m\omega^2 x$ .......

$(I)$

Force applied by the student is

$F= -k \sqrt x$ .......

$(II)$

So, equating

$(I)$ and

$(II)$

$-k \sqrt x = - m\omega^2 x$

$k=m\omega^2 \sqrt x$
so,

$k \propto \sqrt x$

$x$ increases,

$\sqrt x$ increases and thus

$k$ increases.

Hence, the correct answer is option A.

If amplitude of particle executing

$SHM$ is doubled, which of the following quantities are doubled
i) Time period
ii) Maximum velocity
iii) Maximum acceleration
iv) Total energy

0%
ii & iii
0%
i, ii & iii
0%
i, & iii
0%
i, ii, iii & iv

A simple harmonic motion has an amplitude A and time period T. Find the time required by it to travel diameter from .

0%
$x = x\,to\,x = \,A/2$
0%
$x\, = \,0\,\,to\,x\, = \,\frac{A}{{\sqrt 2 }}$
0%
$x = A\,to\,x = \,A/2$
0%
$x = - \frac{A}{{\sqrt 2 }}\,to\,x = \,\frac{A}{{\sqrt 2 }}$

A simple motion is represented by:

$y = 5(\sin 3 \pi t + \sqrt{3} \cos 3\pi t)cm$
The amplitude and time period of the motion are:

0%
$5 \,cm, \dfrac{3}{2}s$
0%
$5 \,cm, \dfrac{2}{3} s$
0%
$10 \,cm, \dfrac{3}{2}s$
0%
$10 \,cm, \dfrac{2}{3}s$

Explanation

$y = 5[\sin (3\pi t) + \sqrt{3} \cos (3\pi t)]$

$= 10 \sin \left(3\pi t + \dfrac{\pi}{3}\right)$

Amplitude

$= 10 \,cm$

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

Which of the following quantity is unitless

0%
Velocity gradient
0%
Pressure gradient
0%
Displacement gradient
0%
NONE

A block of mass m rests on a platform. The platform is given up and down SHM with an amplitude d- What can be the maximum frequency so that the block never leaves the platform?

0%
$\sqrt { \dfrac { g } { d } }$
0%
$\dfrac { 1 } { 2 \pi } \sqrt { \dfrac { g } { d } }$
0%
$\dfrac { 1 } { 2 \pi } \left( \dfrac { g } { d } \right)$
0%
$2 \pi \sqrt { \dfrac { d } { g } }$

Explanation

$\begin{array}{l} m{ w^{ 2 } }d=mg \\ w=\sqrt { \dfrac { g }{ d } } =2\pi t \\ t=\dfrac { 1 }{ { 2\pi } } \sqrt { \dfrac { g }{ d } } \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$

A particle executes simple harmonic motion with an amplitude of 5 cm. when the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. then, its periodic time in second is:

0%
$\dfrac{7}{3}\pi$
0%
$\dfrac{3}{8}\pi$
0%
$\dfrac{4\pi}{3}$
0%
$\dfrac{8\pi}{3}$

Explanation

We know that velocity in S. H. M. is given by

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

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

$a=\omega^2\times 4$

Given |A| = |v|

$4\omega^2=3\omega$

$\omega =\dfrac{3}{4}=\dfrac{2H}{T}$

$T=\dfrac{8H}{3}$ sec.

Two particles undergoing simple harmonic motion of same frequency and same amplitude cross each other at

$x=\dfrac {A}{2}$ . Phase difference between them is

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {2\pi}{3}$
0%
$\dfrac {\pi}{4}$

The amplitude

$(A)$ of damped oscillator becomes half in

$5$ minutes. The amplitude after next

$10$ minutes will be:

0%
$A$
0%
$\cfrac{A}{8}$
0%
$\cfrac{A}{4}$
0%
$4A$

After charging a capacitor

$C$ to a potential

$V$ , it is connected across an ideal inductor

$L$ .The capacitor starts discharging simple harmonically at time

$t = 0 .$ The charge on the capacitor at a later time instant is

$q$ and the periodic time of simple harmonic oscillations is

$T$ . Therefore,

0%
$q = C V \sin ( \omega t )$
0%
$q = C V \cos ( \omega t )$
0%
$T = 2 \pi \sqrt { \frac { 1 } { L C } }$
0%
$T = 2 \pi \sqrt { L C }$

A simple pendulum hanging from the ceiling of a stationary lift has a time period

$T_{1}$ . When the lift moves downward with constant velocity, the time period is

$T_{2}$ , then

0%
$T_{2}$ is infinity
0%
$T_{2}$ > $T_{1}$
0%
$T_{2}$ < $T_{1}$
0%
$T_{2}$ = $T_{1}$

Explanation

$T\propto \dfrac {1}{\sqrt g}$ and

$g$ is same in both cases so time period remain same.

If velocity of a particle in SHM at x=4 m and x=5 m are 15 m/s and 13 m/s then its time period will be:

0%
$\pi$ /4
0%
$\pi$ /2
0%
$\pi$
0%
4 $\pi$ /5

The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:

0%
$\cfrac T 4$
0%
$T$
0%
$\cfrac T 2$
0%
$2 T$

A body oscillates with SHM according to the equation

$x=(5.0\quad m)\cos { [(2\pi \quad rad\quad { s }^{ -1 } } )t+{ \pi }/{ 4] }$
At t=1.5 s, its acceleration is:

0%
$-139.56\quad { m }/{ { s }^{ 2 } }$
0%
$139.56\quad { m }/{ { s }^{ 2 } }$
0%
$69.78\quad { m }/{ { s }^{ 2 } }$
0%
$-69.78\quad { m }/{ { s }^{ 2 } }$

Explanation

$x=5\cos \left(2\pi t+\dfrac{\pi}{4}\right)$

$v=\dfrac{dx}{dt}=-5(2\pi)\sin \left(2\pi t+\dfrac{\pi}{4}\right)$

$a=\dfrac{d^2x}{dt^2}=-5(2\pi)^2\cos \left(2\pi t+\dfrac{\pi}{4}\right)$

Put

$t=1.5$

$a=+139.56m/s^2$

$\cos \left(3\pi +\dfrac{\pi}{4}\right)=\dfrac{-1}{\sqrt{2}}$

So option (B).

1166834_1333712_ans_c1384e564941415288f98f3176655314.jpg

A coin is placed on a horizontal platform, which undergoes horizontal simple harmonic motion about a mean position

$O$ .The coin does not slip on the platform. The force of friction acting on the coin is

$F$ .

0%
$F$ is always directed towards $O$
0%
$F$ is directed towards $O$ when the coin is moving away from $O$ , and away from $O$ when the coin moves towards $O$
0%
$F=0$ when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.
0%
$F$ is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.

A body attached to the lower end of a vertical spring oscillates with time period of 1 sec. The time period when two such springs are connected one below another is approximately:

0%
0.7 sec
0%
1 sec
0%
1.4 sec
0%
2 sec

When the displacement is half of the amplitude, then what fraction of total energy of a simple harmonic oscillator is kinetic:-

0%
$\dfrac {3}{4}th$
0%
$\dfrac {2}{7}th$
0%
$\dfrac {5}{7}th$
0%
$\dfrac {2}{8}th$

a simple harmonic has an amplitude A and time period T. Find the time required by it to travel directly from

$x=0$ to

$\, x=\frac { A}{ \sqrt { 2} }$

0%
T
0%
2T
0%
T/2
0%
T/8

Two bodies A and B of equal mass are suspended from separate massless springs of spring constant

$k_1$ and

$k_2$ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is

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

A simple harmonic wave is represented by the relation

$y\left ( x,t \right )=a_{0}sin2\pi \left ( vt-\dfrac{x}{\lambda } \right )$ if the maximum particle velocity is three times the wave velocity, te wavelength

$\lambda$ of the wave is-

0%
$\pi a_{0}/3$
0%
$2\pi a_{0}/3$
0%
$\pi a_{0}$
0%
$\pi a_{0}/2$

Explanation

$\begin{array}{l} { V_{ p } }={ a_{ 0 } }\left( { 2\pi V } \right) \\ =2\pi { a_{ 0 } }V \\ { V_{ w } }=\dfrac { { 2\pi V\lambda } }{ { 2\pi } } =V\lambda \\ \Rightarrow 2\pi { a_{ 0 } }=3V\lambda \\ \lambda =\dfrac { 2 }{ 3 } \pi { a_{ 0 } } \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$

An object undergoing SHM taken 0.5 s to travel from one point of zero velocity to the next such point. The distance between those point is 50 cm. The period, frequency and amplitude of the motion is:

0%
$1s, 1Hz,25 cm$
0%
$2s.1Hz,50 cm$
0%
$1s, 2Hz,25 cm$
0%
$2s, 2Hz,50 cm$

A particle is executing SHM with time period T Starting from mean position, time taken by it to complete

$\dfrac { 5 }{ 8 }$ oscillations is:

0%
$\dfrac { T }{ 12 }$
0%
$\dfrac { T }{ 6 }$
0%
$\dfrac { 5T }{ 12 }$
0%
$\dfrac { 7T }{ 12 }$

The equation of motion of a particle executing SHM is

$\left( \dfrac { d ^ { 2 } x } { d t ^ { 2 } } \right) + k x = 0$ The time period of the particle will be

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

The function

$sin^{2}(\omega t)$ represents :

0%
a simple harmonic motion with a period $2\pi /\omega$ .
0%
a simple harmonic motion with a period $\pi /\omega$
0%
a periodic, but not simple harmonic motion with period $2\pi /\omega$
0%
a periodic, but not simple harmonic motion with period $\pi /\omega$

A particle performs

$SHM$ on x-axis with amplitude

$A$ and time period

$T$ . The time taken by the particle to a distance

$A/5$ starting from rest is:

0%
$\dfrac { T }{ 20 }$
0%
$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 4 }{ 5 }\right)$
0%
$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 1 }{ 5 }\right)$
0%
$\dfrac { T }{ 2\pi } \sin^{-1} \left(\dfrac { 4 }{ 5 }\right)$

Explanation

Particle is starting from rest, i.e, from one of its extreme position.
As particle moves a distance

$A/5$ , we can represent it on a circle as shown.

$\cos \theta=\dfrac{4A/5}{A}=\dfrac{4}{5}\Rightarrow \theta=\cos^{-1}\left( \dfrac{4}{5}\right)$

$\omega t=\cos^{-1}\left( \dfrac{4}{5}\right)\Rightarrow t=\dfrac{1}{\omega}\cos^{-1}\left( \dfrac{4}{5}\right)=\dfrac{T}{2\pi}\cos^{-1}\left( \dfrac{4}{5}\right)$
1623373_1376259_ans_e3156aeeb5f34f3a8748200447f2842a.PNG

1623373_1376259_ans_e3156aeeb5f34f3a8748200447f2842a.PNG

The figure shows the displacement time graph of a particle executing S.H.M.
If the time period of oscillation is $2 s$ the equation of motion of its SHM

0%
$x = 10 \sin ( \pi t + \pi / 3 )$
0%
$x = 10 \sin \pi t$
0%
$x = 10 \sin ( \pi t + \pi / 6 )$
0%
$x = 10 \sin (2 \pi t + \pi / 6 )$

The displacement

$x$ (in metres) of a particle performing simple harmonic motion is related to time

$t$ (in seconds as}

$x = 0.05 \cos \left( 4 \pi t + \frac { \pi } { 4 } \right)$ . The frequency of the motion will be

0%
$0.5$ $\mathrm { Hz }$
0%
$1.0$ $\mathrm { Hz }$
0%
$1.5$ $\mathrm { Hz }$
0%
$2.0$ $\mathrm { Hz }$

Explanation

Comparing the given equation with the standard equation of SHM i.e.

$y=a\ sin(\omega t+\phi)$

$y=0.05 sin\left(\dfrac{\pi}{2}-(4\pi t+\dfrac{\pi}{4})\right)$

$y=0.05\ sin(\dfrac{\pi}{4}-4\pi t)$

So, the angular frequency of the oscillation will be:

$\omega=4\pi$

So, frequency of oscillation will be:

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

$f=\dfrac{4\pi}{2\pi}=2\ Hz$

Restoring force on the bob of a simple pendulum of mass 100 gm when its amplitude is ${1^\circ}$ :

0%
$1.7 N$
0%
$0.17 N$
0%
$0.017 N$
0%
$0.034 N$

A particle of mass

$0.1 kg$ executes SHM under a force

$F = (-10 x) N$ . Speed of particle at mean position is

$6 m/s$ . Then amplitude of oscillations is

0%
$0.6 m$
0%
$0.2 m$
0%
$0.4 m$
0%
$0.1 m$

Explanation

$\begin{array}{l} M=0.1\, Kg \\ f=-10x \\ \Rightarrow k=10 \\ w=\sqrt { \frac { k }{ m } } \\ =\sqrt { \frac { { 10 } }{ { 0.1 } } } =10 \\ { V_{ \max } }=6m/s \\ \Rightarrow A=\frac { 6 }{ { 10 } } =0.6m \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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