Question

A mass m at the end of a spring oscillates with a frequency of 0.83 Hz.When an additional 730 gmass is added to m, the frequency is 0.65 Hz.What is the value of m?

Answer 1

m will be equal to 1158.73 gram

Step-by-step explanation:

We have given mass m when frequency is 0.83 Hz

So mass
`m_1=m` and frequency
`f_!=f` let spring constant of the spring is KK

Frequency of oscillation of spring is given by
`f=(1)/(2 \pi )\sqrt{(k)/(m)}`

From above relation we can say that
`{(f_1)/(f_2)}=\sqrt{(m_2)/(m_1)}`

It is given that when an additional 730 gram is added to m then frequency become 0.65 Hz ,
`f_2=0.65Hz`

So
`m_2=m+730`

So
`(0.93)/(0.65)=\sqrt{(m+730)/(m)}`

`(m+730)/(m)=1.63`

`0.63m=730`

m= 1158.73 gram