Question

A mass M is attached to an ideal massless spring. When the system is set in motion, it oscillates with a frequency f. What is the new oscillation frequency if the mass is doubled to 2M? g

Answer 1

`\frac {f_i}{\sqrt 2}`

Step-by-step explanation:

Frequency is given by
`\frac {1}{2\pi}\sqrt{{\frac {k}{m}}`

Let the initial frequency be denoted by
`f_i`

`f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}`

When M is 2M then

`f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}`

Therefore, the final frequency is

`\frac {f_i}{\sqrt 2}`