# Questions for Oscillations (JC) Set 1

A mass attached to the lower end of a vertical spring is performing simple harmonic motion. Which of the following statements is correct?

1. When the kinetic energy of the oscillator is equal to its potential energy, the oscillator is neither at the rest position nor at its maximum displacement position.
2. The kinetic energy of the oscillator is a maximum when the tension in the spring is at maximum.
3. The velocity of the mass is zero when its acceleration is zero.
4. The acceleration of the mass is a maximum when it passes through the centre of the oscillation.

Option 2: The kinetic energy of the oscillator is zero when the tension in the spring is at maximum.
Option 3: The velocity of the mass is at a maximum when its acceleration is zero.

Option 4: The acceleration of the mass is zero when it passes through the centre of its oscillation. A particle oscillating in simple harmonic motion has its motion timed at t = 0 s when it is at the 50 cm mark. It travels between the 50 cm and 70 cm marks with a period of 2.0 s. Where is the position of the particle at time t = 0.75 s?

1. 53 cm mark
2. 57 cm mark
3. 63 cm mark
4. 67 cm mark

The equilibrium position is at the 60 cm mark. The +ve is to the right of the 60 cm mark, the -ve is to the left. Since timing begins at the amplitude position (the 50 cm mark),
x = -xocos(2πt/T)
x = – (10)cos[(2π)(0.75)/2)
x = 7.1

x = 7.1 corresponds to the 67.1 cm mark.

A pendulum with a bob of mass m and a string of length L is displaced from its equilibrium position O by a small angle and then released. At the same time, a bob of mass M is dropped and falls vertically downwards through a distance L. A point P is directly below bob of mass M and it happens to be at the same horizontal level as O. The dimensions of the two bobs are the same. Which bob will arrive at its destination first – bob of mass m reaching its equilibrium position O or bob of mass M arriving at the point P?
(Ignore air resistance and you may wish to use: Period of pendulum, T = 2π(L/g)1/2)

1. Bob of mass m.
2. Bob of mass M.
3. The two bobs arrive at their destinations at the same time.
4. The heavier ball will reach its destination first.

The bob of mass m will take tm = T/4 s to reach its equilibrium position O
tm = ¼ T
= ¼(2π(L/g)1/2
= 1.6(L/g)1/2

The bob of mass M experiences free-fall and the time, tM it takes to travel a vertical distance L to arrive at point P is

tM = (2L/g)1/2
= 1.4(L/g)1/2

Since tm > tM, therefore bob of mass M will reach its destination first.

A particle is performing simple harmonic motion between two points A and B. If the period of oscillation is 3.0 s, what is the particle’s maximum acceleration? 1. 4.2 cm s-2
2. 8.8 cm s-2
3. 9.8 cm s-2
4. 18 cm s-2

Use amax = -ω2(-xo), where T = 3s, xo= 2 cm

Back To Oscillations ### 2 thoughts on “Questions for Oscillations (JC) Set 1”

1. how did you get this?
tM = (2L/g)1/2 = 1.4(L/g)1/2

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