MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 1

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

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

0%
$\displaystyle\frac{A}{3}\sqrt{41}$
0%
$\displaystyle 3A$
0%
$\displaystyle A\sqrt 3$
0%
$\displaystyle\frac{7A}{3}$

Explanation

In SHM,

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

$V=\sqrt { { A }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } } =\sqrt { \dfrac { 5{ A }^{ 2 } }{ 9 } }$

$3V=\sqrt { { A' }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } }$

Dividing the above two equations give

$A'=\dfrac { 7A }{ 3 }$

$1\, kg$ block attached to a spring vibrates with a frequency of

$1\,Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in a parallel to an

$8\,kg$ block placed on the same table. So, the frequency of vibration of the

$8\, kg$ block is ;

0%
$\dfrac{1}{4} Hz$
0%
$\dfrac{1}{2 \sqrt{2}} Hz$
0%
$2Hz$
0%
$\dfrac{1}{2} Hz$

Explanation

$f_1=1Hz$

$\displaystyle T_1=2x\sqrt{\dfrac{m_1}{k}}=\frac{1}{f_1}$

$2x=\sqrt{\frac{m}{k}}=1$

$2x\sqrt{\dfrac{1}{k}}=1$

$\dfrac{1}{k}=\left(\frac{1}{2x}\right)^2$

$k=4x^2$

In parallel,

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

$T=2x\sqrt{\dfrac{m_2}{k_{eq}}}=2x\sqrt{\frac{8}{2k}}=2x\sqrt{\dfrac{8}{8x^2}}=2$

$f=\dfrac{1}{2}Hz$

A small block is connected to one end of a massless spring of unstretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t=lt then executes simple harmonic motion with angular frequency

$\omega =\frac { \pi }{ 3 } rad/s$ .
Simultaneously at t=0, a small pebble is projected with speed v from point P at an angle of 45 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is
(take

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

0%
$\sqrt { 50 } m/s$
0%
$\sqrt { 51 } m/s$
0%
$\sqrt { 52 } m/s$
0%
$\sqrt { 53 } m/s$

Explanation

Time is given by

$\displaystyle t=\frac{2vsin \theta}{g}$

$\displaystyle 1= \frac{2vsin45^{o}}{10}$

$\displaystyle v= \sqrt{50}m/s$

A particle is executing SHM along a straight line. Its velocities at distances

$x_1$ and

$x_2$ from the mean position are

$V_1$ and

$V_2$ , respectively. Its time period is:

0%
$2\pi \sqrt {\cfrac {V_1^2+V_2^2}{x_1^2+x_2^2}}$
0%
$2\pi \sqrt {\cfrac {V_1^2-V_2^2}{x_1^2-x_2^2}}$
0%
$2\pi \sqrt {\cfrac {x_1^2+x_2^2}{V_1^2+V_2^2}}$
0%
$2\pi \sqrt {\cfrac {x_2^2-x_1^2}{V_1^2-V_2^2}}$

Explanation

Using the expressions for velocity for SHM,

$v_1 = \omega \sqrt{A^2 -x_1^2}$
$v_2 = \omega \sqrt{A^2 -x_2^2}$
Squaring both,
$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$ ....(3)
$v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)$ ....(4)
Subtracting (4) from (3), $v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2)$
$\Rightarrow 4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$
$\Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}$ where we have used $\omega = \dfrac{2\pi}{T}$

Average velocity of a particle executing SHM in one complete vibration is :

0%
$\dfrac{A \omega}{2}$
0%
$A \omega$
0%
$\dfrac{A \omega^2}{2}$
0%
zero

Explanation

In one complete vibration, displacement is zero. So, average velocity in one complete vibration

$= \dfrac{Displacement}{Time \, interval} = \dfrac{y_f - y_i}{T} = 0$

Hence option D is correct answer

Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is:

0%
$0$
0%
${2\pi}/{3}$
0%
$\pi$
0%
${\pi}/{6}$

Explanation

HINT: DISTANCE IS DEFINED AS THE AREA COVERED BY AN OBJECT FROM MOVING FROM ONE POINT TO ANOTHER

STEP1:In fig (1) particles at a mean position.In fig(2) particles staring some displacement but crossing each other in opposite direction.

STEP2:Now if the displacement of two particle

$\mathrm{x}_{1}=\mathrm{A} \sin (\omega \mathrm{t})$

$\mathrm{x}_{2}=\mathrm{A} \sin (\omega \mathrm{t})$

For $\mathrm{x}_{1}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} \ldots, \dfrac{5 \pi}{6} \ldots, \ldots$

$\mathrm{x}_{2}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} ., \dfrac{5 \pi}{6}$

From fig (2) we can say point $A$ and B corresponding $\omega \mathrm{t}=\dfrac{5 \pi}{6}, \omega \mathrm{t}=\dfrac{\pi}{6}$ So phase difference $\left(\dfrac{5 \pi}{6}-\dfrac{\pi}{6}\right)=\dfrac{2 \pi}{3}$

[N.B $\rightarrow$ Before starting motion of ' 2 '...particle ' 1 ' goes to right extreme end and while '1 returning to mean position particle '2' starts its motion and the point of equal displacements are A and B]

A particle is describing simple harmonic motion. If its velocities are

$v_{1}$ and

$v_{2}$ when the displacements from the mean position are

$y_{1}$ and

$y_{2}$ respectively, then its time period is

0%
$2\pi \sqrt {\dfrac {y_{1}^{2} + y_{2}^{2}}{v_{1}^{2} + v_{2}^{2}}}$
0%
$2\pi \sqrt {\dfrac {v_{2}^{2} - v_{1}^{2}}{y_{1}^{2} - y_{2}^{2}}}$
0%
$2\pi \sqrt {\dfrac {v_{2}^{2} + v_{1}^{2}}{y_{1}^{2} + y_{2}^{2}}}$
0%
$2\pi \sqrt {\dfrac {y_{1}^{2} - y_{2}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

Explanation

In simple harmonic motion,
velocity

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

$\therefore v_{1} = \omega \sqrt {A^{2} - y_{1}^{2}} \Rightarrow v_{1}^{2} = \omega^{2}A^{2} - \omega^{2}y_{1}^{2} .... (i)$
and

$v_{2} = \omega \sqrt {A^{2} - y_{2}^{2}} \Rightarrow v_{2}^{2} = \omega^{2}A^{2} -\omega^{2} y_{2}^{2} ..... (ii)$
Solving equations (i) and (ii), we get

$v_{2}^{2} - v_{1}^{2} = \omega^{2} (y_{1}^{2} - y_{2}^{2})$

$\omega = \sqrt {\dfrac {v_{2}^{2} - v_{1}^{2}}{y_{1}^{2} - y_{2}^{2}}}$

$\Rightarrow T = \dfrac {2\pi}{\omega} = 2\pi \sqrt {\dfrac {y_{1}^{2} - y_{2}^{2}}{v_{1}^{2} - v_{2}^{2}}}$ .

If maximum speed of a particular in SHM is given by

$\displaystyle { V }_{ m }$ . What is its average speed?

0%
$\displaystyle \frac { \pi { V }_{ m }}{2}$
0%
$\displaystyle \frac { 2 V_m}{ \pi }$
0%
$\displaystyle \frac { \pi V_m}{ 4 }$
0%
$\displaystyle \frac { { V }_{ m } }{ \sqrt { 2 } }$

Explanation

Let the SHM be represented as

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

Thus the speed of a particle=

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

$\Rightarrow V_m=A\omega$

The average speed of the particle=

$\dfrac{\int_0^{(2\pi/\omega)}\left|A\omega cos(\omega t+\phi)\right|}{2\pi/\omega}$

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

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

The velocity vector v and displacement vector x of a particle executing SHM are related as

$\dfrac{vdv}{dx}=-\omega^2x$ with the initial condition

$V=v_0$ at

$x=0$ . The velocity v, when displacement is x, is?

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

Explanation

$\begin{array}{l}We\;have\;\\\dfrac{vdv}{dx}=-\omega^2x\;\\or\\\int_{v_0}^vvdv=-\omega^2\int_0^xxdx\\\dfrac{v^2}2-\dfrac{v_0^2}2=-\dfrac{\omega^2x^2}2\\v=\sqrt{v_0^2-\omega^2x^2}\end{array}$

$K_i$ and

$K_p$ respectively are effective spring constant in series and parallel combination of springs as shown in figure, find

$\displaystyle\frac{K_i}{K_p}$

0%
$\displaystyle\frac{9}{2}$
0%
$\displaystyle\frac{3}{7}$
0%
$\displaystyle\frac{2}{9}$
0%
$\displaystyle\frac{7}{3}$

Explanation

In a series combination of above springs, effective spring constant

$K_i=\dfrac{K_1 K_2}{K_1+K_2}$

$\therefore$

$K_i=\dfrac{K\times 2K}{K+2K}$

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

In parallel combination of above springs, effective spring constant

$K_p=K_1+K_2$

$\therefore$

$K_p = K+2K = 3K$

Thus

$\dfrac{K_i}{K_p}=\dfrac{2K/3}{3K}=\dfrac{2}{9}$

In SHM the net force towards mean position is related to its displacement (x) from mean position by the relation

0%
$F \propto x$
0%
$F \propto \frac{1}{x}$
0%
$F \propto x^2$
0%
$F \propto \frac{1}{x^2}$

Explanation

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

$F=-kx$

$\implies F\propto x$

State whether true or false.

Piston of the engine in a motor car exhibits Oscillatory motion.

0%
True
0%
False

The motion described by the needle of sewing machine is oscillatory motion. True or false

0%
True
0%
False

State whether true or false.

All periodic motion need not be oscillatory but all oscillatory motions
are periodic.

0%
True
0%
False

Water waves in a shallow dish are

$6.0\ m$ long. At one point, the water moves up and down at a rate of

$4.8$ oscillations/s. Speed of the wave is:

0%
$28.8\ m/s$
0%
$26.8\ m/s$
0%
$18.8\ m/s$
0%
$38.8\ m/s$

The period of pendulum depends upon

0%
mass
0%
length
0%
amplitude
0%
energy

The relation between T and g by

0%
$T\propto g$
0%
$T\propto g^{2}$
0%
$T^{2}\propto g^{-1}$
0%
$\displaystyle T\propto \frac{1}{g}$

Explanation

T = time period of oscillation

g = effective acceleration on pendulum

l = length of pendulum.

so,

$T=2\pi \sqrt { \dfrac { l }{ g } } \\ \Rightarrow { T }^{ 2 }=4{ \pi }^{ 2 }\dfrac{ l }{ g } \\ \Rightarrow { T }^{ 2 }\propto \dfrac { 1 }{ g }$

When

$l$ is unchanged.

The motion of a pendulum is an example of :

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translatory motion
0%
rotational motion
0%
oscillatotry motion
0%
curvilinear motion

If a body moves back and forth repeatedly about a mean position, it is said to possess

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rotatory motion
0%
projectile motion
0%
oscillatory motion
0%
Reciprocating motion

In the equations below, A, B,

$\omega$ and

$\phi$ are constants;

$y$ and

$t$ are variables;

$t$ represents time. Only one of the following equations does not represent SHM. Which one is that?

0%
$y= A sin \omega t$
0%
$y= B cos \omega t$
0%
$y= A sin \omega t + B cos \omega t$
0%
$y= A sin \omega t + B cos (2\omega t)$

Explanation

A linear combination of coherent simple harmonic motions would result in another simple harmonic motion.

But the combination of two incoherent simple harmonic motions(

$Asin\omega t,B cos(2\omega t)$ ) is NOT an SHM.

Hence the answer is option D.

Displacement-time graph depicting an oscillatory motion is

0%
cos curve
0%
sine curve
0%
tangent curve
0%
straight line

Explanation

The equation for displacement in oscillation is

$x=A\sin(\omega t+\alpha)$

Since

$x$ is a sine function of time, the graph is sinusoidal in nature.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

The time period of a simple pendulum is given by ,

$T=2\pi\sqrt{l/g}$ ,

where

$l=$ distance between point of suspension and center of bob (center of gravity)

It is clear that time period

$T$ of simple pendulum depends on its length

$l$ , hence assertion is incorrect but reason is correct .

One oscillation completed by a vibrating body in one second is known as

0%
1tesla
0%
1Hertz
0%
1horse power.
0%
none

Swinging of table fan is an example of which type of motion?

0%
Oscillatory motion
0%
One dimenetional motion
0%
Projectile motion
0%
None of the above

Which of the following is not an oscillatory motion?

0%
The tuning fork
0%
The stretched string
0%
The motion of the swing
0%
None of the above

If a particle is oscillating on the same horizontal plane on the ground.

0%
It has only kinetic energy but no potential energy
0%
It has only potential energy and no kinetic energy
0%
It has both kinetic and potential energy
0%
None of these

"The projection of the path of a particle in a circle along its diameter is an oscillatory motion" Is this statement correct

0%
This is correct only if diameter of the circle is larger than 2m
0%
This statement is always correct
0%
This statement is always incorrect
0%
This statement is correct only if the particle performs uniform circular motion

In case of force oscillations of a body

0%
driving force is constant throughout.
0%
driving force is to be applied only momentarily.
0%
driving force has to be periodic and continuous.
0%
driving force is not required.

A particle executing simple harmonic motion with an amplitude A. The distance travelled by the particle in one time period is

0%
zero
0%
A
0%
$2$ A
0%
$4$ A

Explanation

The distance travelled by the particle in one time period is

$4$ A.

Time period will be minimum when d is equal to

0%
L/ $\sqrt{12}$
0%
L/ $\sqrt{6}$
0%
L/4
0%
L/2

The equation of motion of a particle is

$x = a cos(\alpha t)^2$ . The motion is

0%
periodic but not oscillatory
0%
periodic and oscillatory
0%
oscillatory but not periodic
0%
neither periodic nor oscillatory.

The function

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

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

What is the effect on the time period of a simple pendulum if the mass of the bob is doubled?

0%
Halved
0%
Doubled
0%
Becomes $8$ times
0%
No effect

Explanation

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

It can be seen that T is independent of mass.

The equation x =a sin 2t + b cos 2t will represent an SHM

0%
True
0%
False

Explanation

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

$\dfrac{d^2x}{dt^2} = -4x.$ . Thus, acceleration is proportional to displacement.

Thus, the equation represents an SHM

The equation of simple harmonic wave is given by

$y=5$ sin

$\dfrac{5}{2}(100 t - x)$ , where

$x$ and

$y$ are in meter and time is in second. The time period of the wave (in second) will be.

0%
$0.04$
0%
$0.06$
0%
$0.01$
0%
$0.05$

Explanation

$y =5$ sin

$\dfrac{\pi}{2}(100 t - x)$

$y= A$ sin

$(\omega t - kx)$

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

$T=\dfrac{2 \pi}{100}$

$\Rightarrow T=0.06$
Hence, the correct option is

$(B)$

The displacement of a particle from its mean position (in meter) is given by

$Y= 0.2\ sin (10 \pi t+1.5 )\ cos\ (10 \pi t+1.5 )$ .
The motion of the particle is

0%
periodic but not $SHM$
0%
Non - Periodic
0%
$SHM$ with period $0.1\ sec$
0%
$SHM$ with period $0.2\ sec$

Equation of SHM is x = 10 sin 10

$\pi$ t. Then the time period is, x is in cm and t is in sec

0%
10 $\pi$
0%
0.2 sec
0%
0.1 sec
0%
2 sec

Explanation

Given,

$x=10\sin(10\pi t)$

Comparing with

$x=A\sin(\omega t)$

$\omega=10\pi$

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{10\pi}=0.2$

Option

$\textbf B$ is the correct answer

For a body in

$S.H.M$ the velocity is given by the relation

$v=\sqrt{144-16x^2}m/sec$ . The maximum acceleration is

0%
$12 m/sec^2$
0%
$16 m/sec^2$
0%
$36 m/sec^2$
0%
$48 m/sec^2$

Explanation

velocity of the particle in SHM,

$v=\sqrt{144-16x^{2}}=4\sqrt{9-x^{2}}$

comparing with

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

we get

$A=3m$

$w=4rad/s$

so, maximum acceleration,

$a_{max}=Aw^{2}$

$=3\times 4^{2}$

$\Rightarrow a_{max}=48m/s^{2}$

The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

0%
$\sqrt{\frac{K}{m}}$
0%
$\sqrt{\frac{m}{K}}$
0%
$\left ( \frac{K}{m} \right )^{2}$
0%
$\left ( \frac{m}{K} \right )^{2}$

Explanation

Suppose you displace the particle by a distance

$'x'$

The spring now exerts a force,

This provides nccenary force for

$SHM$

$\Rightarrow \ F=mwe^2x=k2$ (

$w:$ natural angular frequency )

$\Rightarrow \ w=\sqrt {K/m}$

When a spring having spring constant

$2Nm^{-1}$ is stretched by 5 cm, the energy stored in it is:

0%
0.025 J
0%
0.0025 J
0%
0.25 J
0%
0.5 J

Two particle execute S H M of same amplitude and frequency along the same straight line, They pass one another going in opposite direction each time their displacement is half of their amplitude. The phase difference between them is

0%
$0 ^ { \circ }$
0%
$120 ^ { \circ }$
0%
$180 ^ { \circ }$
0%
$135 ^ { \circ }$

Explanation

$\begin{array}{l} According\, to\, \, question............. \\ In\, first\, case: \\ \, \, \, \, \, \, \frac { a }{ 2 } =a\, sin\, \omega t \\ \Rightarrow sin\, \omega t=\frac { 1 }{ 2 } \\ and\, in\, \, { { second } }\, \, case: \\ \, \, \, \, \, \, \frac { a }{ 2 } =a\, sin\, (\omega t+\varphi ) \\ \Rightarrow sin\, (\omega t+\varphi )=\frac { 1 }{ 2 } \\ Now, \\ \Rightarrow sin\, \omega t\cos \varphi +\cos \omega t\sin \phi =\frac { 1 }{ 2 } \\ \Rightarrow \frac { 1 }{ 2 } \cos \varphi +\sqrt { 1-\frac { 1 }{ 4 } } \sin \varphi =\frac { 1 }{ 2 } \\ \Rightarrow \sqrt { 3 } \sin \phi =1-\cos \varphi \\ \Rightarrow { \sin ^{ 2 } }\varphi =1+{ \cos ^{ 2 } }\varphi -2\cos \varphi \\ \Rightarrow 4{ \cos ^{ 2 } }\varphi -2\cos \varphi -2=0 \\ \Rightarrow \cos \varphi =\frac { { 1-3 } }{ 4 } =-\frac { 1 }{ 2 } \\ \therefore \, \, \, \, \, \cos \varphi =-\frac { 1 }{ 2 } \\ \, \, \, \, \, \, \, \, \, \, \varphi ={ 120^{ 0 } } \\ so\, \, that\, \, the\, correct\, \, option\, is\, B. \end{array}$

The displacement x of a particle in motion is given in terms of time by x (x - 4) = 1 - 5

$\cos \omega t$

0%
The particle executes SHM
0%
The particle executes oscillatory motion which is not SHM
0%
motion of the particle is neither oscillator; nor simple harmonic
0%
The particle is not acted upon by a force when it is at x = 4

Explanation

$\begin{array}{l} x\left( { x-u } \right) =1-5\cos wt \\ { x^{ 2 } }-4x=1-5\cos wt \\ x=\dfrac { { 4\pm \sqrt { 16+4\left( { 1-5\cos wt } \right) } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 20-2\cos wt } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 20\left( { 2{ { \sin }^{ 2 } } } \right) wt } } }{ 2 } \\ =\dfrac { { 4\pm \sqrt { 40 } \sin wt } }{ 2 } \\ =2\pm \sqrt { 10 } \sin wt \\ So.\, particle\, perform\, SHM \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$

A particle of mass

$3$ kg, attached to a spring with force constant

$48$ N/m execute simple harmonic motion on a frictionless horizontal surface. The time period of oscillation of the particle, in seconds, is?

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$2\pi$
0%
$8\pi$
0%
$\dfrac{\pi}{8}$

Explanation

The time period of mass,

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

Given,

$m=3$ kg

$k=48$ N/m

$T=2\pi\sqrt{\dfrac{3}{48}}=2\pi\sqrt{\dfrac{1}{16}}$

$=2\pi\times \dfrac{1}{4}=\dfrac{\pi}{2}$ .

Function

$x = A \sin ^ { 2 } \omega t + B \cos ^ { 2 } \omega t + C \sin \omega t \cos \omega t$ represents SHM

0%
For any value of A ,B and C (except C=0)
0%
$A = - B , C = 2 B , \text { amplitude } = | B \sqrt { 2 } |$
0%
$A = B ; C = 0$
0%
$A = B ; C = 2 B , \text { amplitude } = | B |$

Explanation

$\begin{array}{l} x=A{ \sin ^{ 2 } }\omega t+B{ \cos ^{ 2 } }\omega t+C\sin \omega \cos \omega t \\ =\dfrac { { A\left( { 1-\cos 2\omega t } \right) } }{ 2 } +\dfrac { { B\left( { \cos 2\omega t-1 } \right) } }{ 2 } +\dfrac { C }{ 2 } \sin 2\omega t \\ =\left( { \dfrac { A }{ 2 } -\dfrac { B }{ 2 } } \right) +\left( { \dfrac { B }{ 2 } -\dfrac { A }{ 2 } } \right) \cos \omega t+\dfrac { c }{ 2 } \sin \omega t \\ if\, \, \, \, A=B\, \, \, \, \, \, \, \& \, \, \, C=2B \\ x=B+B\sin 2\omega t \\ \therefore Amplitude\, \, =B \end{array}$

$\therefore$ Option

$D$ is correct.

For a simple pendulum, graph between

$L$ and

$T$ will be a :

0%
hyperbola
0%
parabola
0%
straight line
0%
a curve line

Explanation

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

$\Rightarrow l\propto T^{2}$ (Equation of parabola)

The phase difference between the displacement and acceleration of particle executing S.H.M. in radian

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

Explanation

SHM is represented by

$x=A\sin\omega t$

Then, the velocity and acceleration after differentiation will be

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

$a=-A\omega^2\sin\omega t$

Now, we see displacement is in

$\sin\omega t$ & velocity is in

$\cos\omega t$

$\Rightarrow$ Phase difference between x &

$v=\pi/2$ rad

Similarly, we see velocity in

$\cos\omega t$ & acceleration in

$\sin \omega t$

$\Rightarrow$ Phase difference between v &

$a=\pi/2$ rad

Thus, we can conclude that the phase difference between x &

$a=\pi$ rad.

1179750_1209200_ans_33a23fa72d354acabb0abc64078e2bb5.jpg

A particle of mass

$2 kg$ moves along x - axis with potential energy depending upon 'x' co-ordinate as

$U_(x) = (x^2 - 2 x ) J$ . The mass is

$2 kg$ .

0%
SHM with period $\pi$ second
0%
SHM with period $2 \pi$ second
0%
oscillatory but not SHM
0%
SHM with period $\frac{1}{2\pi}$ second

The diagram shows a ball hanging on a string. The ball swings from point

$W$ to point

$Z$ and back to point

$W$ .
Which statement about the ball is correct?

0%
The kinetic energy of the ball is greatest at point $W$
0%
The kinetic energy of the ball is greatest at point $X$
0%
The kinetic energy of the ball is greatest at point $Y$
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The kinetic energy of the ball is the same at all points of the swing

Explanation

It is a case of simple pendulum .

we know that

by conservation of energy

$KE+PE=$ constant.

At point Y , PE is min.

So, the kinetic energy at mean position is maximum in this case.

Since ,Y is the mean position , so, the kinetic energy is max. at point

$Y$ .

Displacement vs, time curve for a particle executing S.H.M. is shown in Figure. Choose the correct statements.

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Phase of the oscillator is same at $t = 0 s$ and $t = 2 s$
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Phase of the oscillator is same at $t = 2 s$ and $t =6 s$
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Phase of the oscillator is same at $t = 1 s$ and $t = 7 s$
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Phase of the oscillator is same at $t =1 s$ and $t = 5 s$ .

Explanation

(b, d) : Two particles are said to be in same phase if the mode of vibration is same i.e., their distance will be

$n \lambda (/n = 1, 2, 3,....)$

The motion of a pendulum is

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rotatory
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oscillatory
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curvilinear
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rectilinear

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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