Oscillations Test

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Oscillations Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

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

A
Amplitude

B
Frequency

C
Period

D
Position of block

E
Total mechanical energy of the block

Solution

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.
Question 2
1 / -0

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:

A

B

C

D

Solution

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.
As $$\omega^2$$ 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.
Question 3
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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.

A
II only

B
I and II only

C
I and III only

D
II and III only

E
I, II and III

Solution

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.
Question 4
1 / -0

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

A
$$6:1$$

B
$$12:1$$

C
$$1:6$$

D
$$1:12$$

Solution

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 $$w_h = \dfrac{2\pi}{43200} = \dfrac{\pi}{21600}$$ $$rad/s$$
$$\therefore$$ $$\dfrac{w_m}{w_h} =\dfrac{\dfrac{\pi}{1800}}{\dfrac{\pi}{21600}} = 12$$
$$\implies$$ $$w_m : w_h =12:1$$
Question 5
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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 $$g=10 m/s^2$$.

A
0.02 m

B
0.14m

C
0.0004m

D
0.2m

E
42.52m

Solution

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, $$mg=kx$$
$$\implies x=\dfrac{mg}{k}$$
$$=\dfrac{36\times 10}{15000}$$
$$\approx 0.02m$$
Question 6
1 / -0

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)?

A
$$x=6\,sin(10t+\dfrac{1}{2}\pi)$$

B
$$x=6\,sin(10\pi t)$$

C
$$x=6\,sin(10\pi t-\dfrac{1}{2}\pi)$$

D
$$x=6\,sin(10t-\dfrac{1}{4}\pi)$$

Solution

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.
Question 7
1 / -0

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

A
Movements is always up and down.

B
Movement is always side to side.

C
All SHM systems are man-made.

D
All SHM systems involve springs.

E
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.

Solution

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.
Question 8
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When a spring-mass system vibrates with simple harmonic motion, the mass in motion reaches its maximum velocity:

A
when its acceleration is greatest

B
when its acceleration is least.

C
once during one oscillation.

D
when it is as its maximum displacement from equilibrium.

E
when it experiences maximum force.

Solution

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 .
Question 9
1 / -0

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?

A
A

B
B

C
C

D
Both A and C

E
Both A and B

Solution

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 .
Question 10
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The black graph pictured below represents the poition -time graph for a spring -mass system oscillating withnsimple harmonic motion.
The colored, dashed graphs represents shapes of possible velocity time graphs for the same motion.
The vertical axis stands for position or velocity, but the scaling does not matter. The time axis is the same for all graphs.
Which colored graph best represents the possible velocity for the mass in this spring

A
red

B
blue

C
green

D
The velocity graph is the same as the position graph

E
All three colored graphs are equally possible.

Solution

For a spring mass system(an SHM), the position of mass is given by
$$x=Asin(\omega t+\phi)$$
The velocity of the mass would be $$v=\dfrac{dx}{dt}=A\omega cos(\omega t+\phi)$$
Hence the functions representing them must be $$90^{\circ}$$ out of phase. Only the green curve can represent the velocity-time graph.