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

Answer:

Explanation:

Given

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

Frequency in SHM is given by

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

where k=spring constant

m=mass of object

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

$k'=0.5 k$

$f'=\frac{1}{2\pi }\sqrt{\frac{0.5 k}{m}}$

$f'=\frac{1}{\sqrt{2}}\times \frac{1}{2\pi }\sqrt{\frac{k}{m}}$

i.e. $f'=\frac{1}{\sqrt{2}}\times 1$

$f'=\frac{1}{\sqrt{2}}$

$f'=0.707\ Hz$