(A) The cord vibrates in its simplest standing-wave mode (fundamental mode),where the length of the cord $d$ corresponds to half a wavelength,i.e.,$d

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(A) The cord vibrates in its simplest standing-wave mode (fundamental mode),where the length of the cord $d$ corresponds to half a wavelength,i.e.,$d = \frac{\lambda}{2}$,or $\lambda = 2d$.The frequency of the standing wave $f$ is equal to the frequency of the $AC$ current.The wave speed $v$ on a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$.Using the wave equation $v = f\lambda$,we substitute $\lambda = 2d$ to get $v = f(2d) = 2fd$.Equating the two expressions for $v$: $\sqrt{\frac{T}{\mu}} = 2fd$.Squaring both sides: $\frac{T}{\mu} = 4f^2 d^2$.Therefore,the relationship is $T = 4\mu f^2 d^2$.

Class 11 Physics Waves Transverse Stationary Waves and Sonometer

Difficult

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

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Transverse Stationary Waves and Sonometer

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