Simple Harmonic Motion Test

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Simple Harmonic Motion Test - 23

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Question 1
1 / -0

Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement.

A
D will vibrate with maximum amplitude.

B
C will vibrate with maximum amplitude.

C
B will vibrate with maximum amplitude.

D
All the four will oscillate with equal amplitude.

Solution

Since length of pendulum A and C is same and , hence their time period is same and they will have same frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.
Question 2
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The equation of motion of a particle is x = acos(αt)². The motion is

A
periodic but not oscillatory

B
periodic and oscillatory

C
oscillatory but not periodic

D
neither periodic nor oscillatory.

Solution

The motion is oscillatory but not periodic because the oscillations are followed from the maximum displacement (negative extreme) at t=0, the corresponding echo is equal to π/2
Question 3
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Which of the following statements is/are true for a simple harmonic oscillator?

a. Force acting is directly proportional to displacement from the mean position and opposite to it

b. Motion is periodic

c. Acceleration of the oscillator is constant

d. The velocity is periodic

A
(a, b, d)

B
(a, c)

C
(b, d)

D
(c, d)

Solution
Question 4
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The displacement time graph of a particle executing SHM is shown in the figure. Which of the following statement is/are true?

a) the force is zero at t = 3T/4

b) the acceleration is maximum at t = 4T/4

c) the velocity is maximum at t = T/4

d) the PE is equal to KE of oscillation t = T/2

A
(a, b, d)

B
(a, b, c)

C
(b, c, d)

D
(c, d)

Solution

At mean position (x=0), maximum velocity occurs.

As a result, velocity is maximum between T/4 and 3T/4, not at T/2.

At extreme positions, such as 0, T/2, T, acceleration is maximum. As a result, B is right.

At the mean position, i.e. at T/4, 3T/4, the force or acceleration is zero. As a result, C is right.

When PE=0, which occurs at the mean position, KE=TE.

T/2 is extreme not mean, thus D is incorrect.
Question 5
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A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure.) Take the direction from A to B as the positive direction and choose the correct statements.

AO=OB=5 cm

BC= 8 cm

a. The sign of velocity, acceleration and force on the particle, when it is 3 cm away from A going towards B, are positive.

b. The sign of velocity of the particle at C going towards B is negative.

c. The sign of velocity, acceleration and force on the particle, when it is 4 cm away from B going towards A, are negative.

d. The sign of acceleration and force on the particle when it is at points B is negative.

A
(a, b, d)

B
(a, c, d)

C
(b, c, d)

D
(c, d)

Solution

Hint: The direction and velocity and acceleration of the particle depend on the position of the particle.

Consider the diagram.

Step 1: Find the direction and velocity and acceleration of the particle at different positions.

1. When the particle is 3 cm away from A going towards B, velocity is towards AB i.e., positive.

In SHM, acceleration is always towards the mean position (O) in this case.

Hence, it is positive.

2. When the particle is at C, velocity is towards B, hence positive.

3. When the particle is 4 cm away from B going towards A, velocity is negative and acceleration is towards mean position (O), hence, negative.

4. Acceleration is always towards mean position (O). When the particle is at B, acceleration and force are towards BA that is negative.
Question 6
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The rotation of the earth about its axis is:

a. periodic motion

b. simple harmonic motion

c. periodic but not simple harmonic motion

d. non-periodic motion

A
(a, b, d)

B
(a, c)

C
(b, d)

D
(c, d)

Solution

Hint: In SHM, the body does to and fro motion.

The rotation of the earth about its axis is periodic because it repeats after a regular interval of time.

The rotation of the earth is obviously not a to and fro type of motion about a fixed point, hence its motion is not SHM.
Question 7
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The motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is

a. simple harmonic motion

b. non-periodic motion

c. periodic motion

d. periodic but not SHM

A
(a, b)

B
(a, d)

C
(c, d)

D
(b, c)

Solution

Hint: In a periodic motion, the body repeats its motion after a certain time interval.

The motion of the ball bearing will be an SHM but non-periodic as the ball bearing loses its energy due to friction.
Question 8
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From the given functions, identify the function which represents a periodic motion:

A
e^ωt

B
log_e(ωt)

C
sinωt + cosωt

D
e^−ωt

Solution

Hint: Identify the periodic function

Step 1: Recall the definition of periodic motion

Step 2: Draw a graph for each option

Step 3: Express option (3) as a single function of sine
Question 9
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On average, a human heart is found to beat 75 times in a minute. Its frequency and period respectively are:

A
2.25 Hz and 0.8 sec

B
1.35 Hz and 0.9 sec

C
1.25 Hz and 0.8 sec

D
1.25 kHz and 0.8 sec

Solution

Hint: Frequency is defined as the number of beats per second.

Step 1: Find the beat frequency of the heart.

The beat frequency of heart = 75/(1 min)

= 75/(60 sec)

= 1.25 s^–1

= 1.25 Hz

Step 2: Find the time period of the beating of the heart.

The time period, T = 1/(1.25 s^–1) = 0.8 sec
Question 10
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Which of the following functions of time represent/s periodic motion? [ω is any positive constant]

(a) sinωt + cosωt

(b) sinωt + cos2ωt + sin4ωt

(c) e^–ωt

(d) log(ωt)

(e) sin²ωt

A
(a)

B
(a), (b) and (d)

C
(a), (b) and (e)

D
All

Solution

Hint: A motion that repeats itself at regular intervals of time is called periodic motion.

Step 1: Find the equations which are representing the periodic motion.

i) sinωt + cosωt is a periodic function, it can also be written as

The periodic time of the function is 2π/ω.

(ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency.

(iii) The function e^–ωt is not periodic, it decreases monotonically with increasing time and tends to zero as t → ∞ and thus, never repeats its value.

(iv) The function log(ωt) increases monotonically with time t. It, therefore, never repeats its value and is a non-periodic function. It may be noted that as t

→∞, log (ωt) diverges to ∞. It, therefore, cannot represent any kind of physical displacement.

The function is periodic having a period T = π/ω. It also represents a harmonic motion with the point of equilibrium occurring at 1/2 instead of zero.