Question

If the period of oscillation of mass m suspended from a spring is 2 sec, then the period of mass 4m will be
A. 1sec
B. 2sec
C. 3sec
D. 4sec

Answer 1

Hint: The time period is the time taken to complete one full oscillation. For the suspended mass and spring system, the period of oscillation is 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 increase in the period of oscillation.

Formula used:
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
Where T is the period, k is the spring constant and m is the mass of the block.

Complete step by step solution:
For the initial case;
Let the spring constant of the given spring through which the mass is suspended to form the vertical spring mass oscillator is k. The mass of the block is given as m and the period of oscillator is 2sec. Then using the formula of period of spring-mass oscillator is,
\[T = 2\pi \sqrt {\dfrac{m}{k}} \]
\[2\sec = 2\pi \sqrt {\dfrac{m}{k}} \]

For the final case;
The block is replaced with another block of mass 4m and the spring is same as used in the initial case. So, using the formula of period of spring-mass oscillator, the final period will be;
\[{T_2} = 2\pi \sqrt {\dfrac{{{m_2}}}{{{k_2}}}} \]
\[{T_2} = 2\pi \sqrt {\dfrac{{4m}}{k}} \]
\[{T_2} = 2\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)\]

On dividing the first case period with the second period, we get
\[\dfrac{{2\sec }}{{{T_2}}} = \dfrac{{\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}{{2\left( {2\pi \sqrt {\dfrac{m}{k}} } \right)}}\]
\[\dfrac{{2\sec }}{{{T_2}}} = \dfrac{1}{2}\]
\[{T_2} = 4\sec \]

So, the period of the oscillator becomes 4sec when the mass is increased four times keeping the spring same.

Therefore, the correct option is (D).

Note: We should be careful while using the formula of period of suspended mass-spring system. We should assume that the spring used in the spring-mass system is massless or having negligible mass.