Question

A horizontal spring has one end firmly attached to a wall and the other end attached to a mass. The mass can slide freely on a smooth, horizontal surface with no friction. The mass is pulled away from the equilibrium position by a distance A in the positive x-direction and then released so that it oscillates in simple harmonic motion with frequency f. What will happen to the frequency of the oscillation if the mass is doubled?

Answer 1

Doubling the mass attached to a spring in a frictionless simple harmonic oscillator will decrease the frequency of oscillation by a factor of √2, or approximately 0.7071 times the original frequency.

Step-by-step explanation:

In a simple harmonic oscillator like the mass-spring system described, the frequency of oscillation is given by f = (1/2π) * √(k/m), where k represents the spring constant and m the mass attached to the spring. If the mass is doubled, the frequency of the oscillation will decrease because the frequency is inversely proportional to the square root of the mass. Therefore, if the mass increases by a factor of two, the new frequency will be f' = (1/2π) * √(k/2m), which is f' = f/√2. This implies that the frequency will decrease by a factor of √2, or approximately 0.7071 times the original frequency.

Answer 2

Step-by-step explanation:

The frequency of a spring is:

f = 1/(2pi) sqrt(k / m)

If m doubles, then f decreases by a factor of 1/sqrt(2).