A weight suspended from an ideal spring oscillates up and down with a period T. If the amplitude of the oscillation is doubled, the period will be:
Correct answer: A. T
- A. T
- B. 1
- C. 2T
- D. T/2
Explanation
The period of a simple harmonic oscillator, such as a weight on an ideal spring, is given by the formula T = 2π√(m/k), where m is the mass and k is the spring constant. The formula shows that the period T is independent of the amplitude of oscillation. Therefore, even if the amplitude is doubled, the period remains the same. Consequently, the correct answer is T. Option B is incorrect as periods are not represented by unitless integers. Option C is incorrect as the period does not double with amplitude. Option D, T/2, is incorrect as the period does not halve.
Last updated
About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
Practise Waves
1,281 free Waves MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1 rev min-1 is equal to:
85.95° degree in terms of radian is
A 0.3 kg mass oscillates with amplitude 0.08 m. If its maximum kinetic energy is 0.096 J, what is the angular frequency?
A 0.5 kg mass on a spring (k = 50 N/m) is damped with a damping constant b = 2 kg/s. What is the damping ratio?
A 1 kg mass oscillates with SHM of amplitude 0.05 m and frequency 2 Hz. What is the total mechanical energy?