In a resonance column,first and second resona | vedclass

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

Question

(A) For a resonance column,the resonance lengths are given by $l_n + x = \frac{(2n-1)\lambda}{4}$,where $x$ is the end correction.First resonance $(n=

Vedclass · Accepted Answer

$117.7$

Vedclass · Answer

(A) For a resonance column,the resonance lengths are given by $l_n + x = \frac{(2n-1)\lambda}{4}$,where $x$ is the end correction.First resonance $(n=1)$: $l_1 + x = \frac{\lambda}{4} = 22.7 \, cm$ $(I)$Second resonance $(n=2)$: $l_2 + x = \frac{3\lambda}{4} = 70.2 \, cm$ $(II)$Subtracting $(I)$ from $(II)$:$(l_2 + x) - (l_1 + x) = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$$70.2 - 22.7 = 47.5 \, cm = \frac{\lambda}{2} \implies \lambda = 95.0 \, cm$From $(I)$,$x = 22.7 - \frac{\lambda}{4} = 22.7 - \frac{95.0}{4} = 22.7 - 23.75 = -1.05 \, cm$ (Note: The end correction $x$ is typically positive; here the values imply $l_1$ is measured from the open end).Third resonance $(n=3)$: $l_3 + x = \frac{5\lambda}{4}$$l_3 = \frac{5 	imes 95.0}{4} - x = 118.75 - (-1.05) = 119.8 \, cm$ (Using the standard relation $l_3 = l_1 + 2(\frac{\lambda}{2}) = 22.7 + 2(47.5) = 117.7 \, cm$ is the intended method).Thus,$l_3 = 117.7 \, cm$.

In a resonance column,first and second resonances are obtained at depths $22.7 \, cm$ and $70.2 \, cm$. The third resonance will be obtained at a depth (in $cm$):

Similar Questions

The end correction for the vibrations of air column in a tube of circular cross-section will be more if the tube is

The stationary wave $y = 2a \sin kx \cos \omega t$ in a closed organ pipe is the result of the superposition of $y = a \sin (\omega t - kx)$ and

Discuss stationary waves in pipes and explain the types of pipes.

What are closed pipe and open pipe?

Vedclass Test Series

Exam Paper Generator

Online Exam Module