Q136 / 170 · Manipal (MET) 2020 · Physics

A mass $m$ hanging from a spring is doing simple harmonic motion with frequency $f$. If the mass is increased by 4 times, then frequency will be

A
$2 f$
Correct answer
B
$f / 2$
C
$4 f$
D
$f / 4$

Solution

Frequency for a mass $m$ executing SHM $f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$ ....(i) When the mass is increased by 4 times, then the new frequency $f^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k}{4 m}}=\frac{1}{2} \cdot \frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{f}{2}$

Hence, the correct answer is option: 'B'.

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