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(D) Let $f_0 = \frac{v}{2L}$ be the fundamental frequency of the open pipe.Its second overtone is $3f_0 = \frac{3v}{2L}$.Let $f_c = \frac{v}{4L}$ be t

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$300$

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(D) Let $f_0 = \frac{v}{2L}$ be the fundamental frequency of the open pipe.Its second overtone is $3f_0 = \frac{3v}{2L}$.Let $f_c = \frac{v}{4L}$ be the fundamental frequency of the closed pipe.The frequencies of a closed pipe are given by $(2n-1)f_c$,where $n=1, 2, 3, \dots$.The third overtone corresponds to $n=4$,so $(f_3)_{	ext{closed}} = (2(4)-1)f_c = 7f_c = \frac{7v}{4L}$.According to the problem,$(f_3)_{	ext{closed}} - (3f_0) = 150 \ Hz$.Substituting the expressions: $\frac{7v}{4L} - \frac{3v}{2L} = 150$.$\frac{7v}{4L} - \frac{6v}{4L} = 150$.$\frac{v}{4L} = 150$.Since the fundamental frequency of the open pipe is $f_0 = \frac{v}{2L}$,we have $f_0 = 2 	imes \frac{v}{4L} = 2 	imes 150 = 300 \ Hz$.

When an open pipe is closed from one end,the third overtone of the closed pipe is higher in frequency by $150 \ Hz$ than the second overtone of an open pipe. The fundamental frequency of the open-ended pipe will be (in $Hz$)

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