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.
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 = -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)
- 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 tm = T/4 s to reach its equilibrium position O
tm = ¼ T
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
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?
- 4.2 cm s-2
- 8.8 cm s-2
- 9.8 cm s-2
- 18 cm s-2
Use amax = -ω2(-xo), where T = 3s, xo= 2 cm