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

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

Answer: 1

**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?**

**53 cm mark****57 cm mark****63 cm mark****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 = -x_{o}cos(2πt/T)

x = – (10)cos[(2π)(0.75)/2)

x = 7.1

x = 7.1 corresponds to the 67.1 cm mark.

Answer: 4

**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})**

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

The bob of mass m will take t_{m} = T/4 s to reach its equilibrium position O

t_{m} = 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, t_{M} it takes to travel a vertical distance L to arrive at point P is

t_{M} = (2L/g)^{1/2}

= 1.4(L/g)^{1/2}

Since t_{m} > t_{M}, therefore bob of mass M will reach its destination first.

Answer: 2

**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?**

**4.2 cm s**^{-2}**8.8 cm s**^{-2}**9.8 cm s**^{-2}**18 cm s**^{-2}

Use a

_{max}= -ω

^{2}(-x

_{o}), where T = 3s, x

_{o}= 2 cm

Answer: 2

how did you get this?

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

Since it is just a free fall qn, I used this kinematics eqn: s = ut + (1/2)at^2

Initial velocity = 0, dist travelled: L, Acceleration: g. Hence,

L = (1/2)gt^2.

You can get the final expression by shifting the terms in the above equation.