(B) The distance between two consecutive nodes in a stationary wave is given by $\frac{\lambda}{2}$.Given that the nodes are at $16 \ cm$ and $46 \ cm

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(B) The distance between two consecutive nodes in a stationary wave is given by $\frac{\lambda}{2}$.Given that the nodes are at $16 \ cm$ and $46 \ cm$ from the open end,the distance between them is:$\frac{\lambda}{2} = 46 \ cm - 16 \ cm = 30 \ cm$.Therefore,the wavelength $\lambda = 2 \times 30 \ cm = 60 \ cm = 0.6 \ m$.The speed of sound $v$ is given by the formula $v = f \lambda$,where $f$ is the frequency.Given $f = 500 \ Hz$ and $\lambda = 0.6 \ m$:$v = 500 \times 0.6 = 300 \ m/s$.

Class 11 Physics Waves and Sound Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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An open pipe resonates with a tuning fork of frequency $500 \ Hz$. It is observed that two successive nodes are formed at distances $16 \ cm$ and $46 \ cm$ from the open end. The speed of sound in air in the pipe is ..... $m/s$.

A
$230$
B
$300$
C
$320$
D
$360$

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Waves and Sound

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

$A$ pipe of $30 \text{ cm}$ long and open at both ends produces harmonics. Which harmonic mode of the pipe resonates with a $1.1 \text{ kHz}$ source? (Given: speed of sound in air $v = 330 \text{ ms}^{-1}$)

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It is desired to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by

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In an organ pipe whose one end is at $x = 0$,the pressure is expressed by $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$,where $x$ is in meters and $t$ is in seconds. The organ pipe can be:

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The lengths of two open organ pipes are $l$ and $(l + \Delta l)$ respectively. Neglecting end correction,the frequency of beats between them will be approximately (Here $v$ is the speed of sound).

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$Assertion :$ The fundamental frequency of an open organ pipe increases as the temperature is increased.
$Reason :$ As the temperature increases,the velocity of sound increases more rapidly than the length of the pipe.

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An open pipe resonates with a tuning fork of frequency $500 \ Hz$. It is observed that two successive nodes are formed at distances $16 \ cm$ and $46 \ cm$ from the open end. The speed of sound in air in the pipe is ..... $m/s$.

Similar Questions

$A$ pipe of $30 \text{ cm}$ long and open at both ends produces harmonics. Which harmonic mode of the pipe resonates with a $1.1 \text{ kHz}$ source? (Given: speed of sound in air $v = 330 \text{ ms}^{-1}$)

It is desired to increase the fundamental resonance frequency in a tube which is closed at one end. This can be achieved by

In an organ pipe whose one end is at $x = 0$,the pressure is expressed by $p = p_0 \cos \frac{3\pi x}{2} \sin 300\pi t$,where $x$ is in meters and $t$ is in seconds. The organ pipe can be:

The lengths of two open organ pipes are $l$ and $(l + \Delta l)$ respectively. Neglecting end correction,the frequency of beats between them will be approximately (Here $v$ is the speed of sound).

$Assertion :$ The fundamental frequency of an open organ pipe increases as the temperature is increased.$Reason :$ As the temperature increases,the velocity of sound increases more rapidly than the length of the pipe.

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$Assertion :$ The fundamental frequency of an open organ pipe increases as the temperature is increased.
$Reason :$ As the temperature increases,the velocity of sound increases more rapidly than the length of the pipe.