$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 \;Hz$. The mass of the wire is $3.5 \times 10^{-2} \;kg$ and its linear mass density is $4.0 \times 10^{-2} \;kg \;m^{-1}$. What is
$(a)$ the speed of a transverse wave on the string,and
$(b)$ the tension in the string?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Given:
Mass of the wire,$m = 3.5 \times 10^{-2} \;kg$
Linear mass density,$\mu = 4.0 \times 10^{-2} \;kg \;m^{-1}$
Frequency of vibration,$f = 45 \;Hz$
Length of the wire,$l = \frac{m}{\mu} = \frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}} = 0.875 \;m$
For the fundamental mode of vibration,the length of the wire $l$ is equal to half the wavelength,i.e.,$l = \frac{\lambda}{2}$.
Therefore,$\lambda = 2l = 2 \times 0.875 = 1.75 \;m$.
$(a)$ The speed of the transverse wave $(v)$ is given by:
$v = f \lambda = 45 \times 1.75 = 78.75 \;m/s$.
$(b)$ The tension $(T)$ in the string is given by the relation $v = \sqrt{\frac{T}{\mu}}$,so $T = v^2 \mu$.
$T = (78.75)^2 \times 4.0 \times 10^{-2} = 6201.5625 \times 0.04 = 248.06 \;N$.

Explore More

Similar Questions

The length of a sonometer wire tuned to a frequency of $250 \ Hz$ is $0.60 \ m$. The frequency of the tuning fork with which the vibrating wire will be in tune when the length is made $0.40 \ m$ is .... $Hz$.

Two similar sonometer wires have fundamental frequencies of $500 \, Hz$. They are under the same tension. By what percentage should the tension be increased in one wire so that the two wires produce $5 \, \text{beats/sec}$ (in $\%$)?

$A$ guitar string of length $90\,cm$ vibrates with a fundamental frequency of $120\,Hz.$ The length of the string producing a fundamental frequency of $180\,Hz$ will be $...........cm$.

$A$ string of length $2 \ m$ is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of $500 \ Hz$,then the waves would travel on it with a velocity of ..... $m/s$.

$A$ string is clamped at both ends and it is vibrating in its $4^{th}$ harmonic. The equation of the stationary wave is $Y = 0.3 \sin(0.157 x) \cos(200\pi t)$. The length of the string is ..... $m$ (all quantities are in $SI$ units).

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo