Question

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of the system by a factor of

Answer 1

Answer:

The frequency changes by a factor of  0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

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

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$  ÷  $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$  x  $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27