A metallic wire with tension T and at tempera | vedclass

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

Question

(C) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $

Vedclass · Accepted Answer

$1 	imes 10^{-4} /^{\circ} C$

Vedclass · Answer

(C) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.Since the mass $M$ of the wire is constant,$\mu = \frac{M}{l}$. Substituting this,$n = \frac{1}{2l} \sqrt{\frac{T \cdot l}{M}} = \frac{1}{2} \sqrt{\frac{T}{M \cdot l}}$.Thus,$n \propto \frac{1}{\sqrt{l}}$.Given $n_1 = 1 \ kHz$ at $30^{\circ} C$ and $n_2 = 1.001 \ kHz$ at $10^{\circ} C$,we have $\frac{n_1}{n_2} = \sqrt{\frac{l_2}{l_1}}$.$\frac{1}{1.001} = \sqrt{\frac{l_2}{l_1}} \Rightarrow \frac{l_2}{l_1} = \frac{1}{(1.001)^2} \approx 1 - 2(0.001) = 0.998$.Using the linear expansion formula $l_2 = l_1(1 - \alpha \Delta t)$,where $\Delta t = 30^{\circ} C - 10^{\circ} C = 20^{\circ} C$.$\frac{l_2}{l_1} = 1 - \alpha(20) = 1 - 0.002$.$20 \alpha = 0.002 \Rightarrow \alpha = \frac{0.002}{20} = 0.0001 = 1 	imes 10^{-4} /^{\circ} C$.

$A$ metallic wire with tension $T$ and at temperature $30^{\circ} C$ vibrates with its fundamental frequency of $1 \ kHz$. The same wire with the same tension but at $10^{\circ} C$ temperature vibrates with a fundamental frequency of $1.001 \ kHz$. The coefficient of linear expansion of the wire is

Similar Questions

Which one of the following is not a calcination reaction?

Heartbeat increases due to the stimulation of .......

The correct order of ionisation energy of $C, N, O, F$ is :-

$A$ public park,in the form of a square,has an area of $(100 \pm 0.2) \; m^2$. The side of the park is ............ $m$.

Product $(A)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module