Question

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executed SHM of time period T. If the mass is increased by m, the time period becomes \[\dfrac{{5T}}{3}\]. Then the ratio of \[\dfrac{m}{M}\] is
A. $\dfrac{25}{9}$
B. $\dfrac{16}{9}$
C. $\dfrac{5}{3}$
D. $\dfrac{3}{5}$

Answer 1

Hint:The time period of a simple harmonic oscillation totally depends upon the spring constant of the spring and the mass of the spring. The square of the time period of the oscillation is directly proportional to the mass of the spring and inversely proportional to the spring constant of the spring.

Complete answer:
Here, the mass which is already attached to the spring is shown as M, while the additional mass attached to the spring is denoted by m. The time period of the system with only the mass M is T, and let that when the additional mass is added, the time becomes T’. Thus, here,
\[T' = \dfrac{{5T}}{3}\]. Now, we know that the time period for a simple harmonic motion is given as follows:
\[T = 2\pi \sqrt {\dfrac{M}{k}} \]
Here, T is the time period, M is the mass of the system and k is a constant which is known as the spring constant whose unit is N/m.
Thus, from the above equation, we can conclude that the time period of a simple harmonic motion is directly proportional to the square root of the mass of the system. Thus the ratio of the masses can be found as shown below:
\[
{(\dfrac{{T'}}{T})^2} = \dfrac{{M + m}}{M} \\
\Rightarrow {(\dfrac{5}{3})^2} = \dfrac{{M + m}}{M} \\
\Rightarrow \dfrac{{25}}{9} = 1 + \dfrac{m}{M} \\
\therefore \dfrac{m}{M} = \dfrac{{16}}{9} \\
\]
Thus the ratio of the masses after the addition of the mass to the system is 16/9.

Hence, option B is the correct answer.

Note: Here, we assume that there is no internal friction on the spring and negligible tension on the spring. Also, it is assumed that there is no external force acting on the system such as damping, which decreases the time period of oscillation exponentially and hence affects the formula used, which is the ideal scenario.