The speed of particle B and D are same

The speed of particle A, B, E are maximum

MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 4

Find the length of a simple pendulum such that its time period is

$2\ s$ .

0%
$99.4\ cm$
0%
$89.4\ cm$
0%
$79.4\ cm$
0%
$109.4\ cm$

Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$

$T^2\, =\, 4\pi ^2\, \displaystyle \frac {L}{g}$

$\Rightarrow\, 2^2\, =\, 4\, \times\, 3.14\, \times\, 3.14\, \times\, \displaystyle \frac {L}{9.8}$

$\Rightarrow\, L\, =\, \displaystyle \frac {4\, \times\, 9.8}{4\, \times\, 3.14\, \times\, 3.14}\, m\, =\, 0.994\, m\, =\, 99.4\, cm$

A particle performs SHM with a period

$T$ and amplitude

$a$ . The mean velocity of particle over the time interval during which it travels a distance

$a/2$ from the extreme position is:

0%
$\dfrac{6a}{T}$
0%
$\dfrac{2a}{T}$
0%
$\dfrac{3a}{T}$
0%
None

Explanation

Mean velocity

$\displaystyle =\frac{displacement}{time}$

$\displaystyle =\frac{\left ( a/2 \right )}{\left ( T/6 \right )}=\frac{3a}{T}$

The time period of a simple pendulum is 0.2 sec. What is its frequency of oscillation?

0%
0.5 hz
0%
5 Hz
0%
50Hz
0%
1Hz

Explanation

Frequency is inverse of time period.

$F= \dfrac { 1 }{ 0.2 } =5\ Hz$

You are designing a pendulum clock to have a period of

$1.0\ s$ . How long should the pendulum be ?

0%
$0.25\ m$
0%
$0.50\ m$
0%
$0.25\ cm$
0%
$0.25\ mm$

Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

Putting

$T = 1$ in eqn. (1),

We get

$1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$

$\Rightarrow\, L\, =\, \displaystyle \frac{g}{4\, \pi^2}\, =\, \displaystyle \frac{9.8}{4\, \times\, 3.14\, \times\, 3.14}m\, \Rightarrow\, L\, =\, \displaystyle \frac{9.8}{39.44}m\, =\, 0.2484\, m\, =\, 0.25\, m$ (approx.)

What is to and fro motion of an object called?

0%
Vibratory or oscillatory motion.
0%
Rotatory motion
0%
Linear motion
0%
Curvilinear motion

How does time period (T) of a seconds pendulum vary with length (L) ?

0%
$T\, \propto\, \sqrt L$
0%
$T\, \propto\, L^2$
0%
$T\, \propto\, L$
0%
T does not depend on L

A particle moving on x-axis has potential energy

$U=2-20x+5x^{2}$ Joule along x-axis. The particle is released at

$r=-3$ . The maximum value of

$x$ will be (x is in meter):

0%
$5\ m$
0%
$3\ m$
0%
$7\ m$
0%
$8\ m$

Explanation

$\displaystyle F=-\frac{dU}{dx}=20-10x$

$F=0$
at

$x=0$
So, from

$x=-3$ to

$x=2,$ amplitude is

$5.$ Hence other extreme position will be

$x=2+5=7$

A particle performing SHM takes time equal to T (time period of SHM) in consecutive appearances at a particular point. This point is

0%
An extreme position
0%
The mean position
0%
Between positive extreme and mean position
0%
Between negative extreme and mean position

A desktop toy pendulum swings back and forth once every

$1.0 s$ . How long is this pendulum?

0%
$0.25\, m$
0%
$0.50\, m$
0%
$0.15\, m$
0%
$0.30\, m$

Explanation

$T\, =\, 2 \pi\,\sqrt{\displaystyle \frac{L}{g}}\, \Rightarrow\, T^2\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$ .. (1)

Putting

$T = 1$ in eqn. (1),

We get

$1\, =\, 4 \pi^2\, \times\, \displaystyle \frac{L}{g}$

Which of the following is not a necessary apparatus for the measurement of acceleration due to gravity

$(g)$ by a simple pendulum?

0%
Stop watch
0%
Screw gauge
0%
Spherometer
0%
Scale

Given equation of SHM is

$\displaystyle x=10\sin 10\pi t.$ Find the distance between the two points where speed is

$\displaystyle 50\pi$ cm/sec; x is in cm and t is in seconds.

0%
$5(\sqrt 3 +1)$ cm
0%
20 cm
0%
17.32 cm
0%
8.66 cm

Explanation

$x=10\,\sin10\pi t$

$\Rightarrow\cfrac{dx}{dt}=v=100\pi\cos(10\pi\,t)$

$\Rightarrow 50\pi=100\pi\cos(10\pi\,t)$

$\Rightarrow \cos(10\pi\,t)=\cfrac{1}{2}$

$\Rightarrow 10\pi\,t=\cfrac{\pi}{6},\cfrac{-\pi}{6}$

So,

$t=\cfrac{1}{60}sec$

Now,

$X=10\sin(10\pi\times\cfrac{1}{60})$

$\Rightarrow X=5\sqrt3\,cm$

Also,

$X=10\sin(10\pi(-\cfrac{1}{60}))$

$X=-5\,cm$

$X = 5\ cm$ (distance is always measured in +ve.)

$Separation=[5\sqrt3-(-5)]\,cm$

$\Rightarrow\,Separation=5(\sqrt 3+1)\,cm$

For a particle performing SHM, equation of motion is given as

$\displaystyle \frac{d^{2}x}{dt^{2}}+4x=0$ . Find the time period

0%
$\displaystyle T=\pi$
0%
$\displaystyle T=2\pi$
0%
$\displaystyle T=3\pi$
0%
$\displaystyle T=4\pi$

Explanation

$\displaystyle \frac{d^{2}x}{dt^{2}}=-4x$

$\displaystyle \omega ^{2}=4$

$\displaystyle \omega=2$
Time period

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

The bob of mass

$50 gms$ of a simple pendulum is replaced by another bob of mass

$75 gms$ . The time period of the simple pendulum

0%
Becomes ${75}/{50}$ times the original period
0%
Becomes ${50}/{75}$ times the original period
0%
Becomes $125$ times the original period
0%
Remains unchanged

Explanation

As the periodic time of a pendulum does not depend on the mass of the bob, so on replacing the bob of mass

$50 gm$ by another bob of mass

$75 gm$ the time period of the simple pendulum will remain unchanged.

The magnitude of acceleration of particle executing SHM at the position of maximum displacement is:

0%
zero
0%
minimum
0%
maximum
0%
none of these

Explanation

The acceleration of the particle executing S.H.M is,

$a = -\omega^2 y.$
At maximum displacement (at extreme position),

$y = \pm a$
Hence

$|a_{max}|= \omega^2a$ . i.e., acceleration will be maximum.

A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period (T) of the pendulum as the water flows out?

0%
T decreases first and then increases.
0%
T increases first and then decreases.
0%
T increases throughout.
0%
T does not change.

Explanation

$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
First distance of comfrom suspension point will increase then decrease.

The equilibrium position of the particle is

0%
x = 4
0%
x = 6
0%
x = 2
0%
x = 3

Explanation

F = 8 - 2x or F = -2(x - 4) at equilibrium position F = 0

$\Rightarrow$ x = 4 is equilibrium position
Hence the motion of particle is SHM with force constant 2 and equilibrium position x = 4
Equilibrium position is x = 4

The bob of a second's pendulum is replaced by another bob of double mass. The new time period will be

0%
4 sec
0%
1 sec
0%
2 sec
0%
3 sec

In the given figure

0%
The speed of particle B and D are same
0%
The speed of particle A, B, E are maximum
0%
The particle F has zero speed
0%
All particles have same speed

Explanation

The given wave is a sine wave , in which particles are in SHM , and the velocity of a particle executing SHM is given by ,

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

where

$a=$ amplitude ,

$y=$ displacement of particle from mean position ,

$\omega=$ angular frequency of particle ,

at mean position A , C and E ,

$y=0$

$v=\omega a$ (maximum velocity) ,

at extreme position B and D ,

$y=a$

$v=0$ (minimum velocity) ,

it is clear that the speeds of particle at B and D are same .

If the period of oscillation of a simple pendulum is 4 second and we want to convert it into a second pendulum, then we have to

0%
Make the length of the pendulum one fourth of the previous length
0%
Double the length of the pendulum
0%
Make the length of the pendulum half of the previous length
0%
Double the mass of the bob

Explanation

The time period of a simple pendulum is proportional to the square root of the length i.e.,

$T \infty \sqrt{l}$ . To convert 4 second to 2 we have to make the length of the pendulum one fourth of the previous length.

The frequency of oscillator of the springs as shown in figure will be

0%
$\displaystyle \frac {1}{2\pi} \sqrt {\frac {(k_1+k_2)m}{k_1k_2}}$
0%
$\displaystyle \frac {1}{2\pi} \sqrt {\frac {k_1k_2}{(k_1+k_2)m}}$
0%
$\displaystyle \frac {1}{2\pi}\sqrt {\frac {k}{m}}$
0%
$\displaystyle 2\pi\sqrt {\frac {k}{m}}$

Explanation

$\displaystyle n=\frac {1}{2\pi}\sqrt{\frac {k}{m}}$
As springs are in parallel, total spring constant

$k$ of system of spring

$\displaystyle \frac {1}{k} = \frac {1}{k_1} + \frac {1}{k_2} = \frac {k_1+k_2}{k_1k_2} k = \frac {k_1k_2}{(k_1+k_2)}$

$\displaystyle n=\frac {1}{2\pi} \sqrt {\frac {k_1k_2}{(k_1+k_2)m}}$

All oscillatory motions are ______

0%
periodic
0%
linear
0%
rotatory
0%
curvilinear

State whether true or false.

The needle of a moving sewing machine executes oscillatory motion.

0%
True
0%
False

Which is not a necessary assumption for simple pendulum

0%
in-extensible, light and flexible string.
0%
heavy but small sized sphere (bob).
0%
angular displacement from the mean position is very less.
0%
volume of the bob is negligible.

Which of the following is an example of motion in one direction?

0%
A runner running on a circular track.
0%
A bullet fired from a gun.
0%
A piece of stone thrown horizontally from a tower.
0%
Motion of a wheel.

A pendulum suspended from the ceiling of a train has a time period T when the train is at rest. When the train is accelerating with a uniform acceleration, the time period will

0%
increase
0%
decrease
0%
become infinite
0%
remain unaffected

If a pendulum takes

$4 sec$ to swing in each direction, find the length of the pendulum.

0%
4.2 m
0%
3.97 m
0%
5.2 m
0%
15.9 m

Explanation

Given: The pendulum takes 4sec to swing in each direction, which means it swings in one side of the pendulum for 4 sec.
So, half time period,

$\frac{T}{2}= 4\ sec$

Time period of simple pendulum,

$T= 2\times 4=8 s$

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

$\Rightarrow 8 =2\pi\sqrt{\dfrac{l}{10}}$

$\Rightarrow l =\frac{64\times 10}{4\times \pi^2}$

$\Rightarrow l\approx 15.9\ m$

Ratio of kinetic energy at mean position to potential energy at

$A/2$ of a particle performing SHM.

0%
$2:1$
0%
$4:1$
0%
$8:1$
0%
$1:1$

Explanation

Kinetic energy

$K=\displaystyle\frac{1}{2}m\omega^2(A^2-y^2)$
At mean position

$y=0$

$K=\displaystyle\frac{1}{2}m\omega^2(A^2)$ .......

$(i)$
Potential energy

$U=\displaystyle\frac{1}{2}m\omega^2y^2$

$U$ at

$y=\displaystyle\frac{A}{2}$

$U=\displaystyle\frac{1}{2}\frac{mA^2}{4}\omega^2$ ........

$(ii)$
Dividing Eq

$(i)$ by Eq

$(ii)$ , we get

$\displaystyle\frac{KE}{U}=\frac{4}{1}$

A particle is executing simple harmonic motion with an amplitude

$A$ and time period

$T$ . The displacement of the particles after

$2 T$ period from its initial position is

0%
$A$
0%
$4A$
0%
$8A$
0%
Zero

A simple pendulum performs simple harmonic motion about

$x = 0$ with an amplitude

$'a'$ and time period

$'T'$ . The speed of the pendulum at

$x = \dfrac{a}{2}$ will be:

0%
$\dfrac {\pi a}{T}$
0%
$\dfrac {3\pi^{2}a}{T}$
0%
$\dfrac {\pi a\sqrt {3}}{T}$
0%
$\dfrac {\pi a\sqrt {3}}{2T}$

Explanation

As simple pendulum performs simple harmonic motion.

$\therefore velocity, v = \omega \sqrt {a^{2} - x^{2}}$
At,

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

$v = \dfrac {2\pi}{T} \sqrt {a^{2} -\left (\dfrac {a}{2}\right )^{2}} = \dfrac {2\pi}{T} \dfrac {\sqrt {3a^{2}}}{2} = \dfrac {\pi a\sqrt {3}}{T}$

Find out the Time period of simple harmonic oscillator vibrating with frequency

$2.5$ Hz and an amplitude of

$0.05$ m:

0%
$0.4 sec$
0%
$0.2 sec$
0%
$8 sec$
0%
$20 sec$
0%
$50 sec$

Explanation

We know that frequency

$f=\dfrac{1}{T}$

So, time period ,

$T=\dfrac{1}{f}=\dfrac{1}{2.5}=0.4 sec$

The option A will be correct.

The displacement time graph of a particle executing S.H.M. is as shown in the figure. The corresponding force-time graph of the particle is:

Explanation

Hint:

According to the Simple harmonic equation of motion, the force is directly proportional to the negative of displacement.

Step 1: Use the Simple Harmonic Equation of Motion,

The simple harmonic equation of motion is defined as,

$F+\omega^2y=0$

$\Rightarrow$

$F=-\omega^2y$ .................(1)

Step 2: Analysing the force and displacement relation.

$\omega^2$ is always positive, Therefore the force is directly proportional to the negative displacement. Hence, the force-time graph will be invert of the displacement-time graph.

Hence, option (C) is the correct answer.

Choose the correct option which describe the simple harmonic motion.

I. The acceleration is constant.

II. The restoring force is proportional to the displacement.

III. The frequency is independent of the amplitude.

0%
II only
0%
I and II only
0%
I and III only
0%
II and III only
0%
I, II and III

Explanation

For a SHM,

$F=ma=-kx$ or

$a=-kx/m$

Acceleration is depends on position so it is not constant and hence I is false.

Here II and III are the fundamental characteristic of the simple harmonic motion. Thus, II and III are only correct.

The ratio of the angular speed of minutes hand and hour hand of a watch is:

0%
$6:1$
0%
$12:1$
0%
$1:6$
0%
$1:12$

Explanation

Time taken by minute hand to rotate by

$2\pi$ radian angle is

$60$ minute

$= 3600$

$s$

$\implies$ Angular speed of minute hand

$w_m = \dfrac{2\pi}{3600} = \dfrac{\pi}{1800}$

$rad/s$

Time taken by hour hand to rotate by

$2\pi$ radian angle is

$12$ hrs

$= 43200$

$s$

$\implies$ Angular speed of minute hand

$w_h = \dfrac{2\pi}{43200} = \dfrac{\pi}{21600}$

$rad/s$

$\therefore$

$\dfrac{w_m}{w_h} =\dfrac{\dfrac{\pi}{1800}}{\dfrac{\pi}{21600}} = 12$

$\implies$

$w_m : w_h =12:1$

The equation of a simple harmonic progressive wave is given by y = A sin (100

$\pi t\, -\, 3x$ ). Find the distance between 2 particles having a phase difference of

$\displaystyle \frac {\pi}{3}$ .

0%
$\displaystyle \frac {\pi}{9} m$
0%
$\displaystyle \frac {\pi}{18} m$
0%
$\displaystyle \frac {\pi}{6} m$
0%
$\displaystyle \frac {\pi}{3} m$

Explanation

The equation of the wave form

$y=A sin(\omega t-kx)$ has wavelength of value

$\lambda=\dfrac{2\pi}{k}=\dfrac{2\pi}{3}$

Distance between two points having phase difference

$\phi=x_2-x_1=\dfrac{\lambda}{2\pi}\times \phi$

$=\dfrac{2\pi/3}{2\pi}\times \dfrac{\pi}{3}m=\dfrac{\pi}{9}m$

For the block under SHM shown in the figure, which of the following quantities is directly proportional to the maximum speed of the block?

0%
Amplitude
0%
Frequency
0%
Period
0%
Position of block
0%
Total mechanical energy of the block

Explanation

The maximum potential energy of a spring is

$U_{max}=\dfrac{1}{2}kA^2$ where k be the force constant and A be the amplitude of the oscillation.

When the block passes through the equilibrium position, the all potential energy is converted into kinetic energy and at this point the speed will be maximum.

Thus,

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv_{max}^2$

so,

$A \propto v_{max}$

A mass of

$36$ kg is kept vertically on the top of a massless spring. What is the maximum compression of the spring if the spring constant is

$15000$ N/m. Assume

$g=10 m/s^2$ .

0%
0.02 m
0%
0.14m
0%
0.0004m
0%
0.2m
0%
42.52m

Explanation

When the maximum compression of the spring occurs, the forces of gravity and that from spring due to compression balance each other.

At that point,

$mg=kx$

$\implies x=\dfrac{mg}{k}$

$=\dfrac{36\times 10}{15000}$

$\approx 0.02m$

A block of mass

$m$ attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where :

0%
the speed is the maximum
0%
the potential energy is the minimum
0%
the speed is the minimum
0%
the restoring force is the minimum
0%
the kinetic energy is the maximum

Explanation

The magnitude of acceleration of the simple harmonic motion is

$a=F/m=kx/m$

So, the acceleration is maximum when the displacement

$x$ from equilibrium will be maximum. Thus, speed will be minimum and also the potential energy will also be maximum and kinetic energy will be minimum at the end points of the oscillation.

For the block under SHM shown in the figure, which of the following quantities varies sinusoidally with time?

0%
Amplitude
0%
Frequency
0%
Period
0%
Position of block
0%
Total mechanical energy of the block

Explanation

For a SHM, the position of block at time is given by ,

$y=Asin(wt+\phi)$

From this we can simply say that the position of block will vary sinusoidally with time t.

A block of mass

$m=4$ kg undergoes simple harmonic motion with amplitude

$A=6$ cm on the frictionless surface. Block is attached to a spring of force constant

$k=400 N/m$ . If the block is at

$x = 6$ cm at time

$t = 0$ and equilibrium position is at

$x=0$ then the blocks position as a function of time (with

$x$ in centimetres and

$t$ in seconds)?

0%
$x=6\,sin(10t+\dfrac{1}{2}\pi)$
0%
$x=6\,sin(10\pi t)$
0%
$x=6\,sin(10\pi t-\dfrac{1}{2}\pi)$
0%
$x=6\,sin(10t-\dfrac{1}{4}\pi)$

Explanation

The position of block executing SHM is $x=A\sin\left(\dfrac{{2}{\pi}t}{T}+{\phi}\right)$

where $\phi$ is the initial phase

given at $t=0$ , $x=6cm$ and $A=6cm$

therefore by above equation $6=6\sin\left(0+{\phi}\right)$

or $1=\sin\phi$

or $\sin\dfrac{\pi}{2}=\sin\phi$

or $\phi=\dfrac{\pi}{2}$

now time period of system $T=2\pi\sqrt\frac{m}{k}$

$T=2\pi\sqrt\frac{4}{400}$

$T=\pi/5$

so position of block $x=6\sin\left(10t+\dfrac{\pi}{2}\right)$

as the initial phase is in positive x-direction therefore $+$ sign is taken there.

Which of the following is true of all systems displaying simple harmonic motion?

0%
Movements is always up and down.
0%
Movement is always side to side.
0%
All SHM systems are man-made.
0%
All SHM systems involve springs.
0%
For all SHM systems, the amount of force restoring an object to its equilibrium position is proportional to the amount of the object's displacement from equilibrium.

Explanation

The basic necessity for a motion to be called a simple harmonic motion is that the resistive force acting on the object is proportional to the object's displacement from equilibrium position.

Hence

$F\propto -x$

$\implies F=-kx$ hence option E is correct.

A mass on a frictionless surface is attached to a spring. The spring is compressed from its equilibrium position, B , to point A , a distance x from B . Point C is also a distance x from B, but in the opposite direction. When the mass is released and allowed to oscillated freely, at what point or points is its velocity maximized?

0%
A
0%
B
0%
C
0%
Both A and C
0%
Both A and B

Explanation

When the mass is released , it will undergo SHM , and velocity

$v$ in SHM is given by ,

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

where

$\omega=$ angular frequency (fixed) ,

$a=$ amplitude (fixed) ,

$y=$ displacement from mean position ,

Now from the above relation we can see that

$v$ will be maximum when

$y$ is minimum i.e.

$y=0$ , which is the mean position of SHM . Therefore velocity is maximum at mean position .

The figure above shows the same ideal spring in three different situations.
First without a mass hanging the spring is unstretched.
Next with the blue mass hanging the spring stretches.
Finally with the red mass hanging the spring stretches twice as much as it did with the blue mass.
How much does the red mass weigh compared to the blue mass?

0%
The red mass weighs twice as much as the blue mass.
0%
The red mass weighs four times as much as the blue mass.
0%
The red mass weighs half as much as the blue mass.
0%
The red mass weighs about 1.4 as much as the blue mass.
0%
We cannot determine how much the red mass weighs compared to the blue mass, because we do not know the unstretched length of the spring.

Explanation

In equilibrium position the restoring force set up in the spring is equal to the weight of block i.e.

$ky=mg$ ,

where

$y=$ stretch in spring ,

$k=$ spring constant ,

now as both the springs are not moving in any direction so they are in equilibrium position ,

for blue mass

$m_{B}g=ky_{B}$ ,

for red mass

$m_{R}g=ky_{R}$ , we are using same springs in both positions so k will be constant ,

but given that ,

$2y_{B}=y_{R}$ ,

$m_{R}g=k\times2y_{B}=2ky_{B}$

therefore

$m_{R}/m_{B}=2$ ,

$m_{R}=2m_{B}$

When the mass passes through B, it has

0%
Kinetic energy but no potential energy
0%
Potential energy but no kinetic energy
0%
Kinetic energy and potential energy
0%
No kinetic or potential energy
0%
Total energy equal to zero

A block is attached to an ideal spring undergoes simple harmonic oscillations of amplitude A. Maximum speed of block is calculated at the end of the spring. If the block is replaced by one with twice the mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will

0%
decrease by a factor of 4
0%
decrease by a factor of 2
0%
decrease by a factor of $\sqrt 2$
0%
remain the same
0%
increase by a factor of 2

Explanation

Maximum speed of the block

$V_{max} = wA$ where

$w = \sqrt{\dfrac{k}{m}}$

$\implies$

$V_{max} = \sqrt{\dfrac{k}{m}} A$ where

$k = constant$

Now the mass is doubled i.e

$m' = 2m$ keeping

$A =constant$

$\therefore$

$V'_{max} = \sqrt{\dfrac{k}{2m}} A = \dfrac{ V_{max}}{\sqrt{2}}$

Thus speed is decreased by a factor of

$\sqrt{2}$ .

Simple harmonic oscillation of a given system can be specified completely by stating its:

0%
amplitude, frequency and initial phase.
0%
amplitude, frequency and wavelength
0%
frequency and wavelength.
0%
frequency, wavelength and initial phase.

Simple harmonic motion (SHM) is a technical term used to describe a certain kind of idealized oscillation. A simple harmonic oscillation has

0%
Fixed frequency and fixed amplitude
0%
Fixed frequency and variable amplitude
0%
Variable frequency and fixed amplitude
0%
Variable frequency and variable amplitude

A spring-mass system oscillates back and forth in positive and negative directions. The graph above shows the acceleration of the mass.
Which of the following graphs best shows the velocity of the mass?
The time axes are scaled the same on all graphs

Explanation

The displacement

$y$ in SHM is represented as ,

$y=a \sin\omega t$ ,

where

$a=$ amplitude ,

$\omega=$ angular frequency ,

therefore velocity ,

$v=dy/dt=a\omega\cos\omega t$

$v=a\omega\sin(\omega t+\pi/2)$ ,

and acceleration ,

$A=dv/dt=-a\omega^{2}\sin(\omega t)=a\omega^{2}\sin(\omega t+\pi)$

it is clear that velocity v is behind by a phase angle of

$\pi/2$ ,which is represented by graph C .

Consider a thingy hanging from a spring. The system is set vibrating by pulling the thingy down below its equilibrium position and then letting it go from rest.The frequency of the oscillation is determined by

0%
the amount of the initial displacement
0%
the mass of the thingy and the properties of the spring
0%
the local gravitational field, $g$
0%
all of the above.

Explanation

The frequency is given by

$f=\dfrac { 1 }{ T } =\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { k }{ m } }$

T = time period of the oscillation

m= mass of the thingy

k = spring constant of the spring

So,it is clearly seen that frequency depends on properties of spring and mass thingy.

We are still looking at the oscillating thingy hanging from a spring. The system was set vibrating by pulling the thingy down below its equilibrium position and then letting it go from rest. If the initial displacement is doubled what happens to the maximum kinetic energy of the thing?

0%
It is unchanged.
0%
It is doubled.
0%
It is increased by a factor of 4.
0%
We can't tell from the information provided.

Explanation

Maximum kinetic energy (KE) occurs when the thingy's speed is greatest and is proportional to the square of that speed. The maximum speed occurring at the equilibrium point is given by

$A\omega$ , A being the amplitude and

$\omega$ being the angular frequency. Since,

$\omega$ is irrespective of displacement so, velocity depends on the amplitude and doubling A results in increasing maximum kinetic energy by a factor of 4.

When a spring-mass system vibrates with simple harmonic motion, the mass in motion reaches its maximum velocity:

0%
when its acceleration is greatest
0%
when its acceleration is least.
0%
once during one oscillation.
0%
when it is as its maximum displacement from equilibrium.
0%
when it experiences maximum force.

Explanation

Velocity in SHM is given by ,

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

where

$\omega=$ angular velocity ,

$a=$ amplitude ,

$y=$ displacement from mean position ,

velocity

$v$ will be maximum when RHS of this relation is maximum , this is maximum when

$y$ is minimum i.e.

$y=0$ ,

now acceleration in SHM is given by ,

$A=-\omega^{2}y$ ,

therefore acceleration at

$y=0$ will be ,

$A=0$ ,

it implies that velocity is maximum when acceleration is least .

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.