Question

A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation

Answer 1

Answer:

Explanation:

The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .

1/2 k x² = mgx

.5 x k x .33² = m x 9.8 x .33

k / m = 59.4

frequency of oscillation =  $\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }$

= $\frac{1}{2\pi} \times\sqrt{59.4}$

= 1.22 per second .