(A) The fundamental frequency $(n_0)$ of a string of length $d$ fixed at both ends is given by the formula: $n_0 = \frac{1}{2d} \sqrt{\frac{T}{\mu}}$.

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(A) The fundamental frequency $(n_0)$ of a string of length $d$ fixed at both ends is given by the formula: $n_0 = \frac{1}{2d} \sqrt{\frac{T}{\mu}}$.Since the cord vibrates in its simplest standing-wave state (fundamental mode),the frequency of vibration $f$ is equal to the fundamental frequency $n_0$.Therefore,$f = \frac{1}{2d} \sqrt{\frac{T}{\mu}}$.Squaring both sides,we get: $f^2 = \frac{1}{4d^2} \cdot \frac{T}{\mu}$.Rearranging to solve for tension $T$: $T = 4\mu f^2 d^2$.

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

Medium

One insulated conductor from a household extension cord has a mass per unit length of $\mu$. $A$ section of this conductor is held under tension between two clamps. $A$ subsection is located in a magnetic field of magnitude $B$ directed perpendicular to the length of the cord. When the cord carries an $AC$ current of $i$ at a frequency of $f$,it vibrates in resonance in its simplest standing-wave vibration state. Determine the relationship that must be satisfied between the separation $d$ of the clamps and the tension $T$ in the cord.

A
$T=4\mu f^2d^2$
B
$T=2\mu f^2d^2$
C
$T=\frac{\mu f^2d^2}{2}$
D
$T=\frac{\mu f^2d^2}{4}$

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Waves and Sound

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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MediumTS EAMCET 2025

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MediumMHT CET 2025

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MediumIIT 2003

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An organ pipe $P_1$ closed at one end vibrating in its first overtone and another pipe $P_2$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of $P_1$ and $P_2$ is

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