In a resonance tube,the first resonance with | vedclass

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

Question

(A) In a resonance tube (which acts as a closed pipe),the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.The

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(A) In a resonance tube (which acts as a closed pipe),the resonance lengths are given by $l_1 = \frac{\lambda}{4}$ and $l_2 = \frac{3\lambda}{4}$.The difference between the two resonance lengths is $l_2 - l_1 = \frac{\lambda}{2}$.Given $l_1 = 16 \ cm = 0.16 \ m$ and $l_2 = 49 \ cm = 0.49 \ m$.Therefore,$\frac{\lambda}{2} = 0.49 \ m - 0.16 \ m = 0.33 \ m$.This implies $\lambda = 0.66 \ m$.Using the wave equation $v = n\lambda$,where $v = 330 \ m/s$ is the velocity of sound and $n$ is the frequency:$n = \frac{v}{\lambda} = \frac{330}{0.66} = 500 \ Hz$.

In a resonance tube,the first resonance with a tuning fork occurs at $16 \ cm$ and the second at $49 \ cm$. If the velocity of sound is $330 \ m/s$,the frequency of the tuning fork is: (in $Hz$)

Similar Questions

The pitch of an organ pipe is highest when the pipe is filled with

If $v$ is the speed of sound in air,then what is the shortest length of a closed pipe that resonates at a frequency $n$?

An organ pipe open at one end is vibrating in its first overtone and is in resonance with another pipe open at both ends vibrating in its third harmonic. The ratio of the length of the two pipes is:

The lengths of the two organ pipes open at both ends are $L$ and $(L+L_1)$. If they are sounded together,the beat frequency will be ($v=$ velocity of sound in air).

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. The frequency of vibration of the air column is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module