A resonance tube completely filled with water | vedclass

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

Question

(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.First,calculate the wavelength $\lambda$ of the so

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given: Length of the tube $L = 0.8 \ m$,Frequency $f = 500 \ Hz$,Speed of sound $v = 340 \ m/s$.First,calculate the wavelength $\lambda$ of the sound wave:$\lambda = \frac{v}{f} = \frac{340}{500} = 0.68 \ m$.The resonance condition for a tube closed at one end is $L_n = \frac{(2n-1)\lambda}{4}$,where $n = 1, 2, 3, \dots$.We need to find the number of values of $n$ such that $L_n \le 0.8 \ m$.For $n=1$: $L_1 = \frac{\lambda}{4} = \frac{0.68}{4} = 0.17 \ m$.For $n=2$: $L_2 = \frac{3\lambda}{4} = 3 	imes 0.17 = 0.51 \ m$.For $n=3$: $L_3 = \frac{5\lambda}{4} = 5 	imes 0.17 = 0.85 \ m$.Since $L_3 = 0.85 \ m > 0.8 \ m$,the third resonance cannot be formed within the tube length.Therefore,only $2$ resonances will be heard.

Similar Questions

The frequency of the fifth harmonic of a closed pipe is equal to the frequency of the third harmonic of an open pipe. If the length of the open pipe is $72 \ cm$,then the length of the closed pipe is: (in $cm$)

An open pipe of length $L_1$ has a fundamental frequency $n_1$,and a closed pipe of length $L_2$ has a fundamental frequency $n_2$. If they are joined to form a single closed pipe of length $L = L_1 + L_2$,what is the fundamental frequency of the combined pipe?

$A$ $1 \; m$ long (both ends open) organ pipe is kept in a gas that has double the density of air at $STP$. Assuming the speed of sound in air at $STP$ is $300 \; m/s$,the frequency difference between the fundamental and second harmonic of this pipe is . . . . . . $Hz$.

The fundamental frequency of a closed organ pipe of length $20\; cm$ is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is ...... $cm$.

In a resonance pipe, the first and second resonances are obtained at depths $22.7 \,cm$ and $70.2 \,cm$ respectively. What will be the end correction (in $\,cm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module