A standing wave in a pipe with a length L=1.2 | vedclass

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(D) The given wave equation is $y(x, t) = y_0 \sin \left( \frac{2\pi}{L} x \right) \sin \left( \frac{2\pi}{L} x + \frac{\pi}{4} \right)$.Comparing thi

Vedclass · Accepted Answer

The frequency of the fundamental mode of vibrations is $137.5 \,Hz$.

Vedclass · Answer

(D) The given wave equation is $y(x, t) = y_0 \sin \left( \frac{2\pi}{L} x ight) \sin \left( \frac{2\pi}{L} x + \frac{\pi}{4} ight)$.Comparing this with the general form of a standing wave $k = \frac{2\pi}{\lambda} = \frac{2\pi}{L}$,we get $\lambda = L = 1.2 \,m$.For a pipe closed at both ends,the allowed wavelengths are $\lambda_n = \frac{2L}{n}$. For $n=2$,$\lambda = L = 1.2 \,m$,which is consistent with the wave equation.At $x=0$,$y(0, t) = y_0 \sin(0) \sin(\pi/4) = 0$,so there is a node at $x=0$.At $x=L/2$,$y(L/2, t) = y_0 \sin(\pi) \sin(\pi + \pi/4) = 0$,which implies a node at $x=L/2$. However,the statement in option $(c)$ claims an antinode at $x=L/2$,which is incorrect based on the equation,but the question asks for the incorrect statement among the choices provided.The fundamental frequency is $f = \frac{v}{\lambda_{max}} = \frac{v}{2L} = \frac{300}{2 	imes 1.2} = \frac{300}{2.4} = 125 \,Hz$.Since $125 \,Hz 
eq 137.5 \,Hz$,option $(d)$ is clearly incorrect.

$A$ standing wave in a pipe with a length $L=1.2 \,m$ is described by $y(x, t)=y_0 \sin [(2 \pi / L) x] \sin [(2 \pi / L) x+\pi / 4]$. Based on the above information,which one of the following statements is incorrect? (Speed of sound in air is $300 \,ms^{-1}$)

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