An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the X-axis. They are pulled from the equilibrium position in opposite directions along the X-axis by a

Question

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the X-axis. They are pulled from the equilibrium position in opposite directions along the X-axis by a small angular amplitude $θ_{0} = \cos^{- 1} (0.9)$ and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is 330 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity $g = 10 {m/s}^{2}$ ) is _______. [2025]

Smart Achievers · Accepted Answer

(32)

$From Doppler's effect,$

$f_{\max} = \frac{v + v'}{v - v'} f$

$f_{\min} = \frac{v - v'}{v + v'} f$

$Δ f_{\max} = f_{\max} - f_{\min} = \frac{v + v'}{v - v'} f - \frac{v - v'}{v + v'} f$

$= \frac{{(v + v')}^{2} - {(v - v')}^{2}}{v^{2} - v'^{2}} f$

$Δ f_{\max} = \frac{4 v v'}{v^{2} - v'^{2}} f$

$\cos θ_{0} = 1 - \frac{θ_{0}^{2}}{2} = 0.9$

$\frac{θ_{0}^{2}}{2} = 0.1 \Rightarrow θ_{0} = \sqrt{0.2} = \frac{1}{\sqrt{5}}$

$v' = ℓ θ_{0} ω$ $(ω = angular frequency)$

$= ℓ θ_{0} \sqrt{\frac{g}{ℓ}} = θ_{0} \sqrt{g ℓ}$

$v' = \frac{1}{\sqrt{5}} \sqrt{10 \times 8} = 4$

$Putting value of v = 330 m/s, v' = 4 m/s and f = 660 Hz in equation (i)$

$Δ f_{\max} = \frac{4 \times 330 \times 4 \times 660}{330^{2} - 4^{2}} = \frac{16 \times 330 \times 660}{334 \times 326} \approx 32$

Solution

Want Us to Email you About Special Offers & Updates?