The transverse displacement of a string (clam | vedclass

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

Question

(A) The standard equation for a standing wave is $y(x,t) = A \sin(kx) \cos(\omega t)$.Comparing this with the given equation $y(x,t) = 0.6 \sin \left(

Vedclass · Accepted Answer

$648$

Vedclass · Answer

(A) The standard equation for a standing wave is $y(x,t) = A \sin(kx) \cos(\omega t)$.Comparing this with the given equation $y(x,t) = 0.6 \sin \left( \frac{2\pi}{3}x ight) \cos (120\pi t)$,we get angular frequency $\omega = 120\pi \, rad/s$ and wave number $k = \frac{2\pi}{3} \, m^{-1}$.The wave speed $v$ is given by $v = \frac{\omega}{k} = \frac{120\pi}{2\pi/3} = 180 \, m/s$.The linear mass density $\mu$ is $\mu = \frac{M}{L} = \frac{3.0 	imes 10^{-2} \, kg}{1.5 \, m} = 2 	imes 10^{-2} \, kg/m$.The wave speed in a string is related to tension $T$ by $v = \sqrt{\frac{T}{\mu}}$,so $T = \mu v^2$.Substituting the values: $T = (2 	imes 10^{-2} \, kg/m) 	imes (180 \, m/s)^2 = 2 	imes 10^{-2} 	imes 32400 = 648 \, N$.

Similar Questions

$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $25 \ Hz$. The mass of the wire is $2 \ g$ and its linear mass density is $4 \times 10^{-3} \ kg/m$. What is the tension in the string (in $N$)?

The length of the wire shown in the figure between the pulleys is $1.5 \, m$ and its mass is $12.0 \, g$. The frequency of vibration with which the wire vibrates in three loops,forming an antinode at the midpoint of the wire,is: (Given $g = 9.8 \, m/s^2$)

$n$ waves are produced on a string in $1 \ s$. When the radius of the string is doubled,keeping the tension the same,the number of waves produced in $1 \ s$ for the same harmonic will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module