Question

Two waves executing simple harmonic motion travelling in the same direction with same amplitude and frequency are superimposed The resultant amplitude is equal to the $\sqrt{3}$ times of amplitude of individual motions The phase difference between the two motions is _______(degree)

Answer 1

Correct Answer: 60

To find the phase difference between two superimposed waves resulting in a resultant amplitude that is √3 times the amplitude of the individual waves, we can use the formula for the resultant amplitude when two waves interfere. The expression for the resultant amplitude \(A_r\) when two waves of the same amplitude \(A\) and frequency interfere with a phase difference \(\phi\) is given by:

\(A_r = 2A \cos \left(\frac{\phi}{2}\right)\)

Given that \(A_r = \sqrt{3}A\), we can set up the equation:

\(\sqrt{3}A = 2A \cos \left(\frac{\phi}{2}\right)\)

Cancelling \(A\) from both sides, we have:

\(\sqrt{3} = 2 \cos \left(\frac{\phi}{2}\right)\)

Solving for \(\cos \left(\frac{\phi}{2}\right)\):

\(\cos \left(\frac{\phi}{2}\right) = \frac{\sqrt{3}}{2}\)

We know that \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), therefore:

\(\frac{\phi}{2} = \frac{\pi}{6}\)

Solving for \(\phi\), we have:

\(\phi = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\)

Converting \(\frac{\pi}{3}\) radians to degrees, since \(180^\circ\) equals \(\pi\) radians, we get:

\(\phi = \frac{180^\circ}{3} = 60^\circ\)

This calculated phase difference of 60 degrees is within the given range (60, 60), confirming its validity.