The extension in a string,obeying Hooke's law | vedclass

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

Question

(A) The speed of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass

Vedclass · Accepted Answer

$1.22$

Vedclass · Answer

(A) The speed of a transverse wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.According to Hooke's law,the tension $T$ in the string is proportional to the extension $x$,i.e.,$T = kx$.Therefore,the speed $v$ is proportional to $\sqrt{x}$ (since $\mu$ remains constant),so $v \propto \sqrt{x}$.Given the initial extension is $x$ and speed is $v$,and the new extension is $1.5x$ with new speed $v'$,we have:$\frac{v'}{v} = \sqrt{\frac{1.5x}{x}} = \sqrt{1.5} \approx 1.22$.Thus,the new speed is $v' = 1.22v$.

The extension in a string,obeying Hooke's law,is $x$. The speed of sound in the stretched string is $v$. If the extension in the string is increased to $1.5x$,the speed of sound will be (in $, v$)

Similar Questions

The amplitude of a wave represented by the displacement equation $y = \frac{1}{\sqrt{a}}\sin \omega t \pm \frac{1}{\sqrt{b}}\cos \omega t$ will be

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be the position vectors of the vertices $A, B, C$ respectively of $\triangle ABC$. The vector area of $\triangle ABC$ is :

$A$ particle moves in a circle of radius $25 \, cm$ at two revolutions per second. The acceleration of the particle in $m/s^2$ is

An athlete completes one round of a circular track of radius $R$ in $40 \ s$. What will be his displacement at the end of $2 \ min \ 20 \ s$?

If a normal chord at a point $t$ on the parabola $y^2=4ax$ subtends a right angle at the vertex,then $t^2$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module