Question

An object on a spring oscillates in simple harmonic motion with frequency f = 1 Hz. If the spring is exchanged for a new one with a spring constant that is half as large as the previous one, what is the new frequency of oscillation?

Answer 1

Step-by-step explanation:

Given

Frequency of an object in SHM is
`f=1\ Hz`

Frequency in SHM is given by

`f=(1)/(2\pi )\sqrt{(k)/(m)}`

where k=spring constant

m=mass of object

if spring is exchanged such that new spring constant is half of previous one then

`k'=0.5 k`

`f'=(1)/(2\pi )\sqrt{(0.5 k)/(m)}`

`f'=(1)/(√(2))* (1)/(2\pi )\sqrt{(k)/(m)}`

i.e.
`f'=(1)/(√(2))* 1`

`f'=(1)/(√(2))`

`f'=0.707\ Hz`