Question

A simple harmonic oscillator oscillates with frequency f when its amplitude is A. If the amplitude is now doubled to 2A, what is the new frequency?
A) 2f

B) 4f

C) f

D) f/2

E) f/4

Answer 1

Answer:

Explanation:

Given:

Frequency, f1 = f

Amplitude, A2 = 2 × A1

T = 2pi × sqrt(L/g)

Where,

T = Period

= 1/frequency

L = length of the wire in motion

g = acceleration due to gravity

The period (and frequency) of an object in a Simple harmonic motion is dependent on the length and acceleration due to gravity of the wire.

The acceleration does not change the frequency of the motion.

Option C = f

Answer 2

Answer:

(C) f

Explanation:

For a simple harmonic oscillator in the form of a loaded spring, its frequency is given by

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

where k is the spring constant and m is the mass of the load.

It is observed that the amplitude of the spring does not factor into this equation. Hence, the frequency is independent of the amplitude and remains the same.

The same applies to a simple pendulum whose frequency is given by

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

l is the length of the pendulum and g is the acceleration of gravity.

It is observed as well that the amplitude does not appear in the equation.