Question

Two simple harmonic motions with the same amplitude and same frequency acting in the same direction are impressed on a particle. If the resultant amplitude of the particle is twice the amplitude of individual S.H.Ms, the phase difference between the two simple harmonic motions is

• A. \(\frac{2\pi}{\sqrt{3}}\)
• B. \(\frac{\pi}{2}\)
• C. \(\frac{\pi}{4}\)
• D. \(\frac{2\pi}{3}\)
• E. \(\frac{\pi}{3}\)

Hint:

For two equal SHMs: \[ R=2a\cos\frac{\phi}{2} \] Maximum resultant amplitude \(2a\) happens when phase difference is zero.

Answer 1

The Correct Option is D

Step 1: Write equations for the two SHMs

Let the amplitude of each SHM be \(A\) and angular frequency be \(\omega\).

Let the phase difference between them be \(\phi\).

$$ y_1 = A \sin(\omega t) $$ $$ y_2 = A \sin(\omega t + \phi) $$

Step 2: Find resultant amplitude formula

When two SHMs of same frequency and same direction superpose, the resultant amplitude \(R\) is given by:

$$ R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} $$

Here \(A_1 = A_2 = A\), so:

$$ R = \sqrt{A^2 + A^2 + 2A \cdot A \cos\phi} = \sqrt{2A^2 + 2A^2\cos\phi} $$ $$ R = A\sqrt{2(1 + \cos\phi)} $$

Step 3: Use given condition

Given that resultant amplitude equals individual amplitude:

$$ R = A $$ $$ A = A\sqrt{2(1 + \cos\phi)} $$

Divide both sides by \(A\):

$$ 1 = \sqrt{2(1 + \cos\phi)} $$

Squaring both sides:

$$ 1 = 2(1 + \cos\phi) $$ $$ \frac{1}{2} = 1 + \cos\phi $$ $$ \cos\phi = \frac{1}{2} - 1 = -\frac{1}{2} $$

Step 4: Find phase difference

$$ \phi = \cos^{-1}\left(-\frac{1}{2}\right) $$

We know \(\cos\frac{2\pi}{3} = -\frac{1}{2}\)

$$ \phi = \frac{2\pi}{3} $$

Physical Interpretation: For the resultant to have the same amplitude as each component, the two vectors must be at \(120^\circ\) to each other. This forms an equilateral triangle with the two individual amplitudes and the resultant.