Question 246: Oscillations

Question

A mass $m$ is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period $T$. If the mass is increased by $m_0$, the periodic time becomes $\frac{5 \mathrm{~T}}{4}$. The ratio $\frac{\mathrm{m}_0}{\mathrm{M}}$

Accepted Answer

Let
$m$ = original mass
$m_0$ = extra mass added
$T$ = original time period
$T'$ = new time period $= \frac{5T}{4}$
Time period of a mass-spring system:
$ T = 2\pi \sqrt{\frac{m}{k}} $
where $k$ is the spring constant.
With the new mass $m + m_0$:
$ T' = 2\pi \sqrt{\frac{m + m_0}{k}} $
Given:
$ T' = \frac{5T}{4} $
Step 1: Write the ratio $\frac{T'}{T}$
$ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{m + m_0}{k}}}{2\pi \sqrt{\frac{m}{k}}} = \sqrt{\frac{m + m_0}{m}} $
$ \frac{T'}{T} = \sqrt{1 + \frac{m_0}{m}} $
Step 2: Substitute the given value
$ \frac{5}{4} = \sqrt{1 + \frac{m_0}{m}} $
Step 3: Square both sides
$ \left(\frac{5}{4}\right)^2 = 1 + \frac{m_0}{m} $
$ \frac{25}{16} = 1 + \frac{m_0}{m} $
Step 4: Solve for $\frac{m_0}{m}$
$ \frac{25}{16} - 1 = \frac{m_0}{m} $
$ \frac{25 - 16}{16} = \frac{m_0}{m} $
$ \frac{9}{16} = \frac{m_0}{m} $
Step 5: Answer
Ratio $\frac{m_0}{m} = \frac{9}{16}$
Correct option: Option C

Answer

$\frac{3}{4}$

Answer

$\frac{4}{3}$

Answer

$\frac{9}{16}$

Answer

$\frac{16}{9}$

Oscillations

A mass m is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that mass executes S.H.M. of time period T. If the mass is increased by m_0, the periodic time becomes \frac{5 \mathrm{~T}}{4}. The ratio \frac{\mathrm{m}_0}{\mathrm{M}}

Choose the correct answer: