A pipe of length 1.5 m closed at one end is f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For an open pipe,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.Given $v = 360\ m/s$ and $l = 1.5\ m$,we have $n_0 = \frac{360}{2 	im

Vedclass · Accepted Answer

$683$

Vedclass · Answer

(B) For an open pipe,the fundamental frequency is given by $n_0 = \frac{v}{2l}$.Given $v = 360\ m/s$ and $l = 1.5\ m$,we have $n_0 = \frac{360}{2 	imes 1.5} = \frac{360}{3} = 120\ Hz$.For a closed pipe,the fundamental frequency is $n_1 = \frac{v'}{4l}$,where $v'$ is the speed of sound in the gas at $30\,^{\circ}C$.Since both pipes resonate with the same tuning fork,$n_1 = n_0 = 120\ Hz$.Thus,$120 = \frac{v'}{4 	imes 1.5} \Rightarrow 120 = \frac{v'}{6} \Rightarrow v' = 720\ m/s$.Using the relation $v \propto \sqrt{T}$,the speed of sound in the gas at $0\,^{\circ}C$ $(v_0)$ is related to the speed at $30\,^{\circ}C$ $(v')$ by $v_0 = v' 	imes \sqrt{\frac{T_0}{T_{30}}}$.Here $T_0 = 273\ K$ and $T_{30} = 273 + 30 = 303\ K$.$v_0 = 720 	imes \sqrt{\frac{273}{303}} = 720 	imes \sqrt{0.90099} \approx 720 	imes 0.9492 \approx 683.4\ m/s$.Rounding to the nearest integer,the speed is $683\ m/s$.

Similar Questions

The two nearest harmonics of a tube closed at one end and open at the other end are $220\, Hz$ and $260\, Hz$. What is the fundamental frequency of the system in $Hz$?

$A$ closed pipe and an open pipe have their first overtone equal in frequency. Then the lengths of these pipes are in the ratio

In the fundamental mode, the time required for a sound wave to reach the closed end of a pipe filled with air is $t$ seconds. What is the frequency of vibration of the air column? ($\lambda$ = wavelength of the wave)

If the length of a closed organ pipe is $1.5 \, m$ and the velocity of sound is $330 \, m/s$,then the frequency for the second note is ... $Hz$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module