Question

An object of mass $m$ is suspended from a spring and it executes S.H.M. with frequency of $v$ . If the mass is increased 4 times, the new velocity will be
(A) $2v$
(B) $\dfrac{v}{2}$
(C) $v$
(D) $\dfrac{v}{4}$

Answer 1

Hint:In simple harmonic motion of a mass suspended from a spring, both period and frequency of the motion depend only on mass $m$and force constant $k$. Therefore, we have to use the relation between frequency $f$, mass $m$and force constant $k$ to solve this problem.

Formula used:
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Where, $f$ is frequency, $k$ is force constant and $m$is mass.

Complete step by step answer:
We know that frequency $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Initially, the mass is $m$ and frequency is $v$.
Let us say ${m_1} = m$ and ${f_1} = v$
$v = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} $
Now, the mass has increased four times.
Therefore, ${m_2} = 4m$
There is no change in the value if the force constant $k$.
Therefore, the new frequency,
$
{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{4m}}} \\
\Rightarrow {f_2} = \dfrac{1}{{\sqrt 4 }} \times \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \\
\Rightarrow {f_2} = \dfrac{1}{2} \times \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \\
$
Now, putting $\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} = v$ , which is its initial frequency
$ \therefore {f_2} = \dfrac{1}{2} \times v = \dfrac{v}{2}$

Therefore, new frequency will be $\dfrac{v}{2}$.Hence, option B is the right choice.

Note:In this question, we have determined the relation between velocity and frequency for this particular case. Similarly, the relation between velocity and period can also be determined as it is the inverse of frequency.