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

Answer:

$\frac {f_i}{\sqrt 2}$

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

$\frac {f_i}{\sqrt 2}$