The frequency of a simple pendulum is n oscillations per minute while that of another is (n $$+ 1 )$$ oscillations per minute. The ratio of the length of the first pendulum to the length of second is-

$$\left( \frac { n } { n + 1 } \right) ^ { 2 }$$

$$\left( n + \frac { 1 } { n } \right) ^ { 2 }$$

$$\left( \frac { n + 1 } { n } \right) ^ { 2 }$$

The equation of a damped simple harmonic motion is $$m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0$$. Then the angular frequency of oscillation is

$$\omega =(\frac{k}{m}-\frac{b^{2}}{4m^{2}})^{1/2}$$

$$\omega =(\frac{k}{m}-\frac{b}{4m})^{1/2}$$

$$\omega =(\frac{k}{m}-\frac{b^{2}}{4m})^{1/2}$$

$$\omega =(\frac{k}{m}-\frac{b^{2}}{4m^{2}})$$

Under the action of a force $$ F = -kx^3$$, the motion of a particle is ($$k$$= a positive constant)

periodic but not simple harmonic

If a simple pendulum with length $$ L $$ and mass of the bob $$ m $$ is vibrating with an amplitude "a', then the maximum tension in the string is

$$ m g\left[1+\left(\frac{a}{L}\right)^{2}\right] $$

$$ \frac{m g}{2}\left[1+2\left(\frac{a}{L}\right)^{2}\right] $$

$$ \frac{m g}{2}\left[1+\left(\frac{a}{L}\right)\right]^{2} $$

A simple pendulum with a bob of mass m swings with angular amplitude of $$60^o$$, when its angular displacement $$30^o$$, find the tension of string

$$\dfrac{mg}{3}{(3\sqrt{3} - 2)}$$

A simple pendulum has a bob suspended by an inextensible thread of length $$1$$ metre from a point $$A$$ of suspension. At the extreme position of oscillation, the thread is suddenly caught by a peg at a point $$B$$ distant $$( 1 / 4 )$$ from $$A$$ and the bob begins to oscillate in the new condition. The change in frequency of oscillation of the pendulum is approximately given by $$\left( g=10{ m }/{ s }^{ { 2 } } \right) .$$

$$\dfrac { 1 } { 4 \sqrt { 10 } }$$ Hz

$$\dfrac { \sqrt { 10 } } { 2 }$$ Hz

$$\dfrac { \sqrt { 10 } } { 3 }$$ Hz

$$\dfrac { 1 } { \sqrt { 10 } }$$ Hz

MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 15

A simple pendulum of length

ℓ

$\ell$ has a bob of mass

m

$\mathrm{m}$ . The work done in projecting the bob
horizontally from the mean position to give it an angular displacement of

60^{\circ}

$60^{\circ}$ is

0%
$\operatorname{mg} \frac{\ell}{2}$
0%
$$

\sqrt{3} \frac{\mathrm{mg{\ell}}{2}

$$
0%
$\frac{m g \ell}{2}(2-\sqrt{3})$
0%
zero

A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m.If at any instant its velocity is 40 cm/sec. The displacement will be (here amplitude is0.5m)

0%
0.06 m
0%
0.3 m
0%
0.01 m
0%
1.0 m

The frequency of a simple pendulum is n oscillations per minute while that of another is (n

+ 1)

$+ 1 )$ oscillations per minute. The ratio of the length of the first pendulum to the length of second is-

0%
$\frac { n } { n + 1 }$
0%
$\left( n + \frac { 1 } { n } \right) ^ { 2 }$
0%
$\left( \frac { n + 1 } { n } \right) ^ { 2 }$
0%
$\left( \frac { n } { n + 1 } \right) ^ { 2 }$

The equation of a damped simple harmonic motion is

m \frac{d^{2} x}{d t^{2}} + b \frac{d x}{d t} + k x = 0

$m\frac{d^{2}x}{dt^{2}}+b\frac{dx}{dt}+kx=0$ . Then the angular frequency of oscillation is

0%
$\omega =(\frac{k}{m}-\frac{b^{2}}{4m^{2}})^{1/2}$
0%
$\omega =(\frac{k}{m}-\frac{b}{4m})^{1/2}$
0%
$\omega =(\frac{k}{m}-\frac{b^{2}}{4m})^{1/2}$
0%
$\omega =(\frac{k}{m}-\frac{b^{2}}{4m^{2}})$

A paramedical staff nurse improvises a second's pendulum (time period 2 s) by fixing one end of a string of length

L

$L$ to a ceiling and the other end to a heavy object of negligible size. Within 60 oscillations of this pendulum, she finds that the pulse of a wounded soldier beats 110 times. A symptom of bradycardia is pulse

< 60

$< 60$ per minute and that of tachycardiais

> 100

$> 100$ per minute. Then the length of the string is nearly symptoms of

0%
I $m ,$ bradycardia
0%
$4 \mathrm { m } ,$ bradycardia
0%
I m, tachycardia
0%
$4 \mathrm { m } ,$ tachycardia

A bob of a simple pendulum executing SHM in air has time period

T_{o}

$T_o$ and in water has time period T. Neglect friction. Relation between T and

T_{o}

$T_o$ if density of bob is

4 / 3 \times 10^{3} k g / m^{3}

$4/3 \times 10^3 kg/m^3$ , is

0%
$T=T_o$
0%
$T=\dfrac{T_0}{2}$
0%
$T=2T_o$
0%
$T=4T_o$

The mass of the bob of a simple pendulum of length L is m. If the bob is left from its horizontal position then the speed of the bob and the tension in the thread at the lowest position of the bob will be respectively :-

0%
$\sqrt{2gL}$ and 3 mg
0%
3 mg and $\sqrt{2gL}$
0%
2 mg and $\sqrt{2gL}$
0%
2 gL and 3 mg

A conical pendulum suspended in a lift, which is at rest, is performing U.C.M. in a horizontal circle with speed v and radius r. Now the lift moves upward with acceleration g/If the radius of the circular motion is kept same, the new speed of the bob is :

0%
v
0%
1.224 v
0%
1.414 v
0%
2 v

Under the action of a force

F = - k x^{3}

$F = -kx^3$ , the motion of a particle is (

k

$k$ = a positive constant)

0%
simple harmonic motion
0%
uniformly accelerated motion
0%
not periodic
0%
periodic but not simple harmonic

If a simple pendulum with length

L

$L$ and mass of the bob

m

$m$ is vibrating with an amplitude "a', then
the maximum tension in the string is

0%
$\frac{m g}{2}$
0%
$\frac{m g}{2}\left[1+2\left(\frac{a}{L}\right)^{2}\right]$
0%
$m g\left[1+\left(\frac{a}{L}\right)^{2}\right]$
0%
$\frac{m g}{2}\left[1+\left(\frac{a}{L}\right)\right]^{2}$

A particle of mass m is in SHM, with velocity

V_{0}

$V_0$ at its lowest position. The maximum height attained by the pendulum will be

0%
$\frac{V_0^2}{2g}$
0%
$\sqrt{V_0 g}$
0%
$2 \sqrt \frac{V_0}{g}$
0%
$\sqrt \frac{V_0^2}{4g}$

A particle executes S.H.M with a period of

8 s

$8s$ . Find the time in which the potential energy is equal to half of the total energy.

0%
$2 \ s$
0%
$4 \ s$
0%
$1 \ s$
0%
$0.5 \ s$

A simple pendulum with a bob of mass m swings with angular amplitude of

60^{o}

$60^o$ , when its angular displacement

30^{o}

$30^o$ , find the tension of string

0%
$\dfrac{mg}{2}$
0%
$\dfrac{3\sqrt{3}mg}{2}$
0%
$\dfrac{mg}{3}{(3\sqrt{3} - 2)}$
0%
$\dfrac{\sqrt{3}mg}{2}$

Two graphs of the same harmonic wave are shown below. The graph (1) on the left shows the displacement of wave y, as a function of position x for a given instant of time. The graph (2) on the right shows the displacement of a wave as a function of time 't' for a given position. The speed of the wave is

0%
5.0 $cms^{2}$
0%
0.5 $cms^{2}$
0%
0.4 $cms^{2}$
0%
4.0 $cms^{2}$

\to F

$\vec { F }$ force vector,

\to v

$\vec { v }$ is velocity vector,

\to a

$\vec { a }$ vector is acceleration vector and

\to r

$\vec { r}$ vector is displacement vector with respect to mean position than which of the following quantities are always non-negative in a simple harmonic motion along a straight line?

0%
$\vec { F } .\vec { a}$
0%
$\vec { V } .\vec { r}$
0%
$\vec { a } .\vec { r}$
0%
$\vec { F } .\vec { r}$

A simple pendulum of length

ℓ

$\ell$ carries a bob of mass

m .

$m.$ When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is

0%
$m g$
0%
$3 m g$
0%
$\sqrt { 10 } { mg }$
0%
$4 m g$

A simple harmonic motion is given by

y = 5 (s i n 3 π t + \sqrt{3} c o s 3 π t)

$y= 5 ( sin 3\pi t +\sqrt3 cos 3 \pi t)$ . What is the amplitude of motion if y in m?

0%
$10$ cm
0%
$5$ cm
0%
$200$ cm
0%
$1000$ cm

A simple pendulum is vibrating with an an angolar amplitude of

90^{\circ}

$90 ^ { \circ }$ ,For what values of with vertical, the accelcration is directed
In sertically upwards
2) horizontally
vertically downwerds

0%
$0 ^ { \circ } , \cos ^ { - 1 } \frac { 1 } { \sqrt { 3 } } , 90 ^ { \circ }$
0%
$\cos ^ { - 4 } \left( \frac { 1 } { \sqrt { 3 } } \right) \cdot \cos w$
0%
$90 ^ { \circ } , \cos ^ { - 1 } \frac { 1 } { \sqrt { 3 } } , 0 ^ { \circ }$
0%
$\cos ^ { 4 } \left( \frac { 1 } { \sqrt { 3 } } \right) \cdot \infty , \sigma$

Two SHMs are represented by the equations

y_{1} = 0.1 s i n

${ y }_{ 1 }=0.1sin$

[100 π + (π / 3)]

$\left[ 100\pi +\left( \pi /3 \right) \right]$ and

y_{2} = 0.1 c o s π t

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

0%
$-\dfrac { \pi }{ 6 }$
0%
$\dfrac { \pi }{ 3 }$
0%
$-\dfrac { \pi }{ 3 }$
0%
$\dfrac { \pi }{ 6 }$

Time period of a simple pendulum is

T

$T$ and its angular amplitude is

θ_{0}

$\theta_{0}$ . Find time taken by the bob to move from equilibrium position

O

$O$ to

P

$P$ if at this instant string makes an angle

θ

$\theta$ with the vertical.

0%
$\dfrac{T}{\pi}\sin^{-1}\dfrac{\theta}{\theta_{0}}$
0%
$\dfrac{T}{2 \pi}\sin^{-1}\dfrac{\theta}{\theta_{0}}$
0%
$T\sin^{-1}\dfrac{\theta_{0}}{\theta}$
0%
$T\sin^{-1}\dfrac{\theta}{\theta_{0}}$

A simple pendulum has a bob suspended by an inextensible thread of length

1

$1$ metre from a point

A

$A$ of suspension. At the extreme position of oscillation, the thread is suddenly caught by a peg at a point

B

$B$ distant

(1 / 4)

$( 1 / 4 )$ from

A

$A$ and the bob begins to oscillate in the new condition. The change in frequency of oscillation of the pendulum is approximately given by

(g = 10 m / s^{2}) .

$\left( g=10{ m }/{ s }^{ { 2 } } \right) .$

0%
$\dfrac { \sqrt { 10 } } { 2 }$ Hz
0%
$\dfrac { 1 } { 4 \sqrt { 10 } }$ Hz
0%
$\dfrac { \sqrt { 10 } } { 3 }$ Hz
0%
$\dfrac { 1 } { \sqrt { 10 } }$ Hz

For a particle executing SHM along xx- axis force is given by:

0%
-Akx
0%
A cos kx
0%
A exp ( -kx)
0%
Akx

A particle executes simple harmonic motion according to equation

$4\dfrac{{{d^2}x}}{{d{t^2}}} + 320 x = 0$ . Its time period of oscillation is

0%
$\dfrac{{2\pi }}{{5\sqrt 3 }}\ s$
0%
$\dfrac{\pi }{{3\sqrt 2 }}\ s$
0%
$\dfrac{\pi }{{2\sqrt 5 }}\ s$
0%
$\dfrac{{2\pi }}{{\sqrt 3 }}\ s$

The displacement of a body executing S.H.M from its mean portion is given by

$x = 0.5 \sin ( 10 \pi t + 1.5 ) \cos ( 10 \pi t + 1.5 )$ . The ratio of the maximum velocity to the maximum acceleration of the body is given by

0%
$20 \pi$
0%
$\frac { 1 } { 20 \pi }$
0%
$\frac { 1 } { 10 \pi }$
0%
$10 \pi$

A particle performs SHM with amplitude A 8time period T. The mean velocity of theparticle over the time interval during which it travels a distance of

$\dfrac{A }{ 2}$ starting fromextreme position is

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

A simple pendulum of period T has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate its period will be

0%
Remains equal to T
0%
Less than T
0%
Greater than T
0%
Infinite

A particle moves under force

$F=-5{(x-2)^3}$ .Motion of the particle is

0%
Translatory
0%
Oscillatory
0%
SHM
0%
All of these

A particle executes SHM. Its instantaneous acceleration is given by a = - px, where p is a constant and x is the displacement from the mean position. The angular frequency of the particle is given by.

0%
$\sqrt{p}$
0%
$1/ \sqrt{p}$
0%
p
0%
1 / p

A wooden cylinder of mass 20 g and area of cross-section 1

$c{m^2}$ , having a piece of lead of mass 60 g attached to its bottom floats in water. The cylinder is depressed and then released.Show that it will execute S.H.M .Find the frequency of oscillations.

0%
0.557 ${s^{ - 1 }}$
0%
0.55 ${s^{ - 1}}$
0%
0.545 ${s^{ - 1}}$
0%
0.513 ${s^{ - 1}}$

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5J If its total energy is 9J and Its amplitude is 0.01 its time period would be

0%
$\frac { \pi } { 10 } \mathrm { sec }$
0%
$\frac { \pi } { 20 } \mathrm { sec }$
0%
$\frac { \pi } { 50 } \mathrm { sec }$
0%
$\frac { \pi } { 100 } \mathrm { sec }$

A particle is performing SHM with energy of vibration

$90$

$\mathrm{J}$ and amplitude 6

$\mathrm{cm}$ . When the particlere aches at distance 4

$\mathrm{cm}$ from mean position, it is stopped for a moment and then released. The new energy of vibration will be

0%
40 $\mathrm{J}$
0%
50 $\mathrm{J}$
0%
90 $\mathrm{J}$
0%
60 $\mathrm{J}$

If the equation of an SHM is

$y =asin(4\pi t+\theta)$ ,how much is its frequency?

0%
2 $Hz$
0%
1/2 $Hz$
0%
$2\pi$ $Hz$
0%
$1/ 2\pi$ $Hz$

If a simple pendulum is taken to a place where g decreases by 2%, then the time period

0%
increases by 1%
0%
decreases by 1%
0%
increases by 2%
0%
decreases by 2%

A particle has displacement y given by y=3 sin

$(5 \pi t + \pi)$ where y is in metre and t is in second. What are frequency and period of motion?

0%
0.4 Hz, 2.5 sec
0%
2.5 Hz, 0.4 sec
0%
2.5 Hz, 2.5 sec
0%
0.4 Hz, 0.4 sec

A small body of mass 0.10 kg is undergoing simple harmonic motion of amplitude 1.0 meter and period of 0.2 sec. The maximum value of the force acting on it is :

0%
99 N
0%
66 N
0%
33 N
0%
11 N

Two pendulums have time periods

$T$ and 5

$T / 4$ They start SHM at the same time from the mean position. After how many oscillations of the small pendulum they will be again in the same phase?

0%
5
0%
4
0%
11
0%
9

In simple harmonic motion, the graph between kinetic energy K and time 't' is :

0%
0%
0%
none
0%
both

If the equation of an SHM is y=a sin

$(4 \pi t + \theta)$ , how much is its frequency?

0%
2
0%
$1/2$
0%
$2 \pi$
0%
$1/2 \pi$

The oscillation of a body on a smooth horizontal surface is represented by the equation,

$X=A\cos{(\omega t)}$
where

$X=$ displacement at time

$T,\omega=$ frequency of oscillation
Which one of the following graphs shows correctly the variation

$a$ with

$t$ ?
Here

$a=$ acceleration at time,

$T=$ time period

A small spherical steel ball is placed a little away from the centre of a large concave mirror whose radius of curvature

$R=2.5cm$ . When the ball is released it begins to oscillate about the centre, the motion of the ball is is simple harmonic then the period of motion is Neglect friction, and take

$g=10m/{sec}^{2}$ .

0%
$1.423$ sec
0%
$2.412$ sec
0%
$3.142$ sec
0%
$3.802$ sec

The reading of a spring balance when a block is suspended from it in air is

$60 N$ . The reading is changed to

$40 N$ when the block is submerged in water. The specific gravity of the block must be therefore.

0%
3/2
0%
6
0%
2
0%
3

The equation of a particle in

$S.H.M$ is

$a+16{\pi}^2x=0$ . In the equation

$'a'$ is the linear acceleration (in

$m/sec^2$ ) of the particle at a displacement

$'x'$ in meter. The time period of

$SHM$ in seconds is :

0%
$\frac{1}{4}$
0%
$\frac{1}{2}$
0%
$1$
0%
$2$

A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum starts swinging in phase. They again swing in phase after an interval of 18 second from the start. The period of the simple pendulum is

0%
0.9 sec
0%
1.8 sec
0%
2.7 sec
0%
3.6 sec

Wave has simple harmonic motion whose period is

$4$ seconds while another wave which also possess simple harmonic motion has its period

$3$ second. If both are combined, then the resultant wave will have the period equal to

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

A wedge of mass m is kept on smooth surface and connected with two springs as shown in the figure.Initially springs are in their natural length. Time period of small oscillations of the wadge will be

0%
$\cfrac {3 \pi} 2 \sqrt{ \cfrac m k}$
0%
$\pi \begin{pmatrix} 1+\cfrac { 1 }{ \sqrt { 3 } } \end{pmatrix}\sqrt { \cfrac { m }{ k } }$
0%
$\pi \begin{pmatrix} 1+\cfrac { 2 }{ \sqrt { 3 } } \end{pmatrix}\sqrt { \cfrac { m }{ k } }$
0%
$\pi \sqrt{ \cfrac m k}$

The phase difference between displacement and acceleration in SHM is

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

Two practice undergo SHM under parallel time in same time period and equal amplitude At particular instant one particle is at extreme position while other is at mean position they mean in same direction they will cross each other is at mean position they mean in in same direction they will cross each other after time

0%
$\dfrac{T}{2}$
0%
$\dfrac{3T}{8}$
0%
$\dfrac{T}{6}$
0%
$\dfrac{3T}{4}$

A mass is considered so that can only move in one dimission along the Xaxis .The potential energy of the mass as s function of X-is shown on the diagram.There is a force in the negative X-direction at.

0%
B
0%
E
0%
C
0%
D

When two simple harmonic motions of same periods, same amplitude, having phase difference of

$3\pi/2$ , and at right angles to each other are super imposed. The resultant wave form is a:

0%
circle
0%
parabola
0%
ellipse
0%
None of these

Two bodies A and B of equal mass are suspended from two spearte massless springs of spring contnat

${ K }_{ 1 }$ and respectively If the bodies oscillate vertically such that their maximum velocities are ewqual , the ratio of the amplitude of A to that of B is

0%
${ K }_{ 1 }/{ K }_{ 2 }\quad \quad \quad$
0%
$\quad \sqrt { { K }_{ 1 }/{ K }_{ 2 } }$
0%
$\\ { K }_{ 2 }/{ K }_{ 1 }\quad$
0%
$\quad \quad \sqrt { { K }_{ 2 }/{ K }_{ 1 } } \quad \quad \quad$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 15 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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