A source of sound S in the form of a block ke | vedclass

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(C) Given, amplitude of spring, $A = 50 \,cm = 0.5 \,m$. Mass of sound source, $m = 100 \,g = 0.1 \,kg$. Speed of sound, $v = 340 \,ms^{-1}$. The maxi

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(C) Given, amplitude of spring, $A = 50 \,cm = 0.5 \,m$. Mass of sound source, $m = 100 \,g = 0.1 \,kg$. Speed of sound, $v = 340 \,ms^{-1}$. The maximum frequency heard by the observer is $n_{\max} = n_{\min} + 0.125 n_{\min} = 1.125 n_{\min}$. Using the Doppler effect formula for a moving source: $n_{\max} = \frac{n_0 v}{v - v_{s\max}} \quad (i)$ $n_{\min} = \frac{n_0 v}{v + v_{s\max}} \quad (ii)$ Dividing $(i)$ by $(ii)$: $\frac{n_{\max}}{n_{\min}} = \frac{v + v_{s\max}}{v - v_{s\max}} = 1.125$ $v + v_{s\max} = 1.125(v - v_{s\max})$ $v + v_{s\max} = 1.125v - 1.125v_{s\max}$ $2.125 v_{s\max} = 0.125v$ $v_{s\max} = \frac{0.125 	imes 340}{2.125} = \frac{42.5}{2.125} = 20 \,ms^{-1}$. For a spring-mass system, $v_{s\max} = A\omega = A\sqrt{\frac{k}{m}}$. $20 = 0.5 \sqrt{\frac{k}{0.1}}$ $40 = \sqrt{\frac{k}{0.1}}$ $1600 = \frac{k}{0.1}$ $k = 160 \,N m^{-1}$. Thus, the correct option is $(c)$.

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