A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system if a mass 4 m is suspended from the same spring?
A. \[\dfrac{n}{4}\]
B. \[4n\]
C. \[\dfrac{n}{2}\]
D. \[2n\]

Hint: The frequency of the spring-mass oscillator is inversely proportional to the square root of the mass of the block attached to the spring. When spring constant remains constant then on increasing the mass of the block will result the decrease in the frequency of oscillation.

Formula used:
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \], here f is the frequency, k is the spring constant and m is the mass of the block.

Complete step by step solution:
The initial mass of the block is given as m. Let the spring constant of the given spring through which the mass is suspended to form the vertical spring mass oscillator is k.

Then using the formula of frequency of spring-mass oscillator is,
\[f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]

The initial frequency is given as n,
So, \[n = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} \]

Now, the initial mass which is suspended to the spring is replaced with another block of mass 4m and the spring is same as used in the initial case.

So, using the formula of frequency of spring-mass oscillator, the final frequency will be;
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{{k_2}}}{{{m_2}}}} \]
\[{f_2} = \dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{{4m}}} \]
\[{f_2} = \dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)\]

On dividing the first frequency with the second frequency, we get
\[\dfrac{n}{{{f_2}}} = \dfrac{{\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}{{\dfrac{1}{2}\left( {\dfrac{1}{{2\pi }}\sqrt {\dfrac{k}{m}} } \right)}}\]
\[\dfrac{n}{{{f_2}}} = 2\]
\[{f_2} = \dfrac{n}{2}\]
So, the frequency of the oscillator becomes \[\dfrac{n}{2}\]when the mass is increased four times keeping the spring same.

Therefore, the correct option is (C).

Note: We should be careful while using the formula of frequency of suspended mass-spring system. If the spring given in the question was not massless than we can’t use that formula as mass of the spring will also contribute to the characteristic of the motion of the system.