Question
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 $$4m$$ is suspended from the same spring ?
A.
$$\frac{n}{4}$$
B.
$$4\,n$$
C.
$$\frac{n}{2}$$
D.
$$2\,n$$
Answer :
$$\frac{n}{2}$$
Solution :
Time period of spring-mass system, is given by
$$\eqalign{
& T = 2\pi \sqrt {\left( {\frac{{{\text{displacement}}}}{{{\text{acceleration}}}}} \right)} \cr
& \therefore {\text{Frequency, }}n = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{{{\text{acceleration}}}}{{{\text{displacement}}}}} \cr
& n = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}} \,......\left( {\text{i}} \right) \cr} $$
In case of vertical spring-mass system, in equilibrium position
$$kl = mg \Rightarrow \frac{g}{l} = \frac{k}{m}$$
where,
$$l =$$ extension in the spring and
$$m =$$ mass of the suspended body
$$k =$$ spring constant or force constant of spring.
$$\therefore $$ From Eq. (i), we have
$$n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \,\,{\text{or}}\,\,n \propto \frac{1}{{\sqrt m }}\,\,{\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} $$
but $${m_1} = m,\,{m_2} = 4m,{n_1} = n\left( {{\text{given}}} \right)$$
$$\therefore \frac{n}{{{n_2}}} = \sqrt {\frac{{4m}}{m}} = 2\,\,{\text{or}}\,\,{n_2} = \frac{n}{2}$$
Alternative
As we know that
$$\eqalign{
& T = 2\pi \sqrt {\frac{m}{k}} \,\,\left( {{\text{for spring mass system}}} \right) \cr
& n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \cr} $$
So, for two different masses suspended with same spring.
$$\eqalign{
& {n_1} = \frac{1}{{2\pi }}\sqrt {\frac{k}{{{m_1}}}} \,\,\left[ {k\,{\text{is}}\,{\text{same}}\,{\text{for}}\,{\text{both}}\,{\text{the}}\,{\text{cases}}\,{\text{as}}\,{\text{spring}}\,{\text{is}}\,{\text{same}}} \right] \cr
& {n_2} = \frac{1}{{2\pi }}\sqrt {\frac{k}{{{m_2}}}} \cr
& {\text{so,}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} \cr
& {\text{here,}}\,\,{m_2} = 4{m_1} \cr
& {\text{so,}}\,\,\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{4{m_1}}}{{{m_1}}}} = \frac{2}{1} \cr
& \Rightarrow {n_1} = 2{n_2} \cr
& \Rightarrow {n_2} = \frac{{{n_1}}}{2} = \frac{n}{2}\,\,\left[ {{n_1} = n} \right] \cr} $$