Question

Two identical sinusoidal waves with wavelengths of 1.5 m travel in the same direction at a speed of 10 m/s. If the two waves originate from the same starting point, but with time delay ∆t between them, and with resultant amplitude A_resultant = √3 A then ∆t will be equal to:

0.00625 sec

0.0125 sec

0.025 sec

0.005 sec

0.01 sec

Clear selection

Answer 1

Let $\Delta t$ be the required time delay or time difference

Phase Difference $\phi = \dfrac{2\pi}{T}\Delta t$

Given,

Velocity of wave v = 10 m/s

Wavelength $\lambda$ = 1.5 m

Now,

$v = f\cdot \lambda\\\Rightarrow f=\dfrac{v}{\lambda}=\dfrac{10}{1.5}=\dfrac{20}{3}$

So, $\dfrac{1}{T}=\dfrac{20}{3}$

Therefore Phase Difference $\phi = \dfrac{2\pi}{T}\Delta t = 2\pi \dfrac{20}{3}\cdot \Delta t=\dfrac{40\pi}{3}\Delta t$

For Resultant Amplitude,

$A_r=\sqrt{A_1^2+A_2^2+2A_1A_2cos\phi}$

In this case $A_1=A_2=A$

So,

$A_r=\sqrt 2A\sqrt{1+cos\phi}\\A_r=2Acos\frac{\phi}{2}$

Given, Resultant amplitude $A_r=\sqrt 3 A$

So,

$\sqrt3 A =2Acos\frac{\phi}{2}\\cos\frac{\phi}{2}=\dfrac{\sqrt 3}{2}\\\dfrac{\phi}{2}=\dfrac{\pi}{6}\\\Rightarrow\phi=\dfrac{\pi}{3}$

$\implies\phi=\dfrac{\pi}{3}=\dfrac{40\pi}{3}\Delta t\\\Rightarrow\Delta t =\dfrac{1}{40}s=0.025\ \ sec$

Answer: $\boxed{\Delta t=0.025\ \ sec}$