Question

A light spring is suspended with mass '\(m_1\)' at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is T second. When a mass '\(m_2\)' is added to '\(m_1\)' and the system is made to oscillate the period is found to be \(\frac{3}{2}\) T. The ratio \((\frac{m_1}{m_2})\) is

• A. \(2:3\)
• B. \(3:4\)
• C. \(4:5\)
• D. \(5:6\)

Hint:

Write each in the form A sin(wt + phi) and compare the phase angles.

Answer 1

The Correct Option is C

Step 1: Rewrite x1:
\(x_1 = \frac{1}{\sqrt2}\sin\omega t + \frac{1}{\sqrt2}\cos\omega t = \sin\omega t\cos\frac\pi4 + \cos\omega t\sin\frac\pi4 = \sin\left(\omega t + \frac\pi4\right)\). Amplitude 1, phase \(\frac\pi4\).

Step 2: Rewrite x2:
\(x_2 = \sin\omega t + \cos\omega t = \sqrt2\left[\frac{1}{\sqrt2}\sin\omega t + \frac{1}{\sqrt2}\cos\omega t\right] = \sqrt2\sin\left(\omega t + \frac\pi4\right)\). Amplitude \(\sqrt2\), phase \(\frac\pi4\).

Step 3: Phase difference:
Both have phase \(\frac\pi4\), so the phase difference is 0. Only the amplitudes differ.

Final Answer:
The phase difference is zero, option (A). \[ \boxed{0} \]