A whistle of frequency 500 , Hz tied to the e | vedclass

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(A) The linear velocity of the whistle is given by $v_S = r\omega$.Given $r = 1.2 \, m$ and angular velocity $\omega = 2\pi 	imes \frac{400}{60} \, r

Vedclass · Accepted Answer

$436 \, Hz$ to $586 \, Hz$

Vedclass · Answer

(A) The linear velocity of the whistle is given by $v_S = r\omega$.Given $r = 1.2 \, m$ and angular velocity $\omega = 2\pi 	imes \frac{400}{60} \, rad/s$.$v_S = 1.2 	imes 2\pi 	imes \frac{400}{60} = 1.2 	imes 2\pi 	imes \frac{20}{3} = 16\pi \approx 50.26 \, m/s$. Using $v_S \approx 50 \, m/s$ as per standard approximation.When the whistle approaches the listener,the observed frequency is maximum: $n_{max} = n \left( \frac{v}{v - v_S} ight) = 500 \left( \frac{340}{340 - 50} ight) = 500 \left( \frac{340}{290} ight) \approx 586 \, Hz$.When the whistle moves away from the listener,the observed frequency is minimum: $n_{min} = n \left( \frac{v}{v + v_S} ight) = 500 \left( \frac{340}{340 + 50} ight) = 500 \left( \frac{340}{390} ight) \approx 436 \, Hz$.Thus,the range of frequencies heard is $436 \, Hz$ to $586 \, Hz$.

$A$ whistle of frequency $500 \, Hz$ tied to the end of a string of length $1.2 \, m$ revolves at $400 \, rev/min$. $A$ listener standing some distance away in the plane of rotation of the whistle hears frequencies in the range (speed of sound $= 340 \, m/s$):

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