Two coherent sources of sound,S_{1} and S_{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the position of the listener be $L$. Initially,the listener is at a distance $x = 2\, m$ from $S_{2}$.The distance from $S_{1}$ is $S_{1}L = \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let the position of the listener be $L$. Initially,the listener is at a distance $x = 2\, m$ from $S_{2}$.The distance from $S_{1}$ is $S_{1}L = \sqrt{x^2 + (1.5)^2} = \sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\, m$.The path difference is $\Delta x = S_{1}L - S_{2}L = 2.5 - 2 = 0.5\, m$.Since $\lambda = 1\, m$,we have $\Delta x = \frac{\lambda}{2}$,which corresponds to a minimum (destructive interference).When the listener moves away from $S_{1}$ while keeping the distance from $S_{2}$ fixed at $2\, m$,the path difference $\Delta x = S_{1}L - S_{2}L$ increases.The next intensity maximum occurs when the path difference $\Delta x = n\lambda$. For the adjacent maximum,$n = 1$.Thus,$\Delta x = 1\, m$.Let $d$ be the distance from $S_{1}$ to the listener at this new position.Then,$d - 2 = 1$,which gives $d = 3\, m$.

Similar Questions

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 the amplitude of individual motions. The phase difference between the two motions is $.....( ^{\circ)}$

Two sound waves each of wavelength $\lambda$ and having the same amplitude $A$ from two sources $S_1$ and $S_2$ interfere at a point $P$. If the path difference $S_2P - S_1P = \lambda/3$,then the amplitude of the resultant wave at point $P$ will be $[\cos(120^{\circ}) = -0.5]$.

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,$y_1(x, t) = 4 \sin(kx - \omega t)$ and $y_2(x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3})$,are (Take the angular frequency of initial waves same as $\omega$):

Two waves of amplitudes $A_1$ and $A_2$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$. The value of $A_2 / A_1$ is [Assume $A_1 > A_2$]

The resultant amplitude due to the superposition of two waves $y_1 = 5 \sin (\omega t - kx)$ and $y_2 = -5 \cos (\omega t - kx - 150^{\circ})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module