Question

A mass attached to a spring oscillates with a period of 22 sec. After 22 kg are​ added, the period becomes 33 sec. Assuming that we can neglect any damping or external​ forces, determine how much mass was originally attached to the spring.

Answer 1

Mass of 17.854 kg is only attached to the spring

Step-by-step explanation:

We have given time period in first case is 22 sec

Let initially mass is m

After 22 kg are added the period becomes 33 sec

Time period of spring mass system is

`T=2\pi \sqrt{(m)/(k)}`, here m is mass and k is spring constant

From the relation we can see that

`(T_1)/(T_2)=\sqrt{(m)/(m+22)}`

`(22)/(33)=\sqrt{(m)/(m+22)}`

Squaring both side

`0.444={(m)/(m+22)}`

`0.444m+9.777=m`

m = 17.584 kg

So mass of 17.854 kg is only attached to the spring