A musician using an open flute of length 50,c | vedclass

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Question

(A) The frequency of the $n^{th}$ harmonic for an open pipe is given by $f_n = \frac{n v_s}{2L}.$ For the second harmonic $(n=2)$,$f = \frac{2 v_s}{2L

Vedclass · Accepted Answer

$666$

Vedclass · Answer

(A) The frequency of the $n^{th}$ harmonic for an open pipe is given by $f_n = \frac{n v_s}{2L}.$ For the second harmonic $(n=2)$,$f = \frac{2 v_s}{2L} = \frac{v_s}{L}.$Given $v_s = 330\,m/s$ and $L = 0.5\,m,$ the source frequency is $f = \frac{330}{0.5} = 660\,Hz.$The person is moving towards the source,so the observed frequency $f'$ is given by the Doppler effect formula: $f' = f \left( \frac{v_s + v_o}{v_s} ight),$where $v_o = 10\,km/h = 10 	imes \frac{5}{18} = \frac{25}{9} \approx 2.78\,m/s.$Substituting the values: $f' = 660 \left( \frac{330 + 2.78}{330} ight) = 660 \left( 1 + \frac{2.78}{330} ight) = 660 + 660 	imes 0.00842 \approx 660 + 5.56 \approx 665.56\,Hz.$Rounding to the nearest integer,the frequency is $666\,Hz.$

$A$ musician using an open flute of length $50\,cm$ produces second harmonic sound waves. $A$ person runs towards the musician from another end of a hall at a speed of $10\,km/h.$ If the wave speed is $330\,m/s,$ the frequency heard by the running person shall be close to...... $Hz$

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