$(i)$ For the wave $y(x, t) = 0.06 \sin \left(\frac{2 \pi}{3} x\right) \cos (120 \pi t)$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The length of the string is $1.5 \, m$ and its mass is $3.0 \times 10^{-2} \, kg$. Do all the points on the string oscillate with the same $(a)$ frequency,$(b)$ phase,$(c)$ amplitude? Explain your answers.
$(ii)$ What is the amplitude of a point $0.375 \, m$ away from one end?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) $(i)$ The given equation represents a standing wave.
$(a)$ Yes,all points on the string oscillate with the same frequency of $f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 \, Hz$,except at the nodes where the amplitude is zero.
$(b)$ No,all points do not oscillate with the same phase. Points within the same loop oscillate in phase,but points in adjacent loops oscillate with a phase difference of $\pi$ radians.
$(c)$ No,all points do not oscillate with the same amplitude. The amplitude of a point at position $x$ is given by $A(x) = 0.06 \sin \left(\frac{2\pi}{3} x\right)$,which depends on the position $x$.
$(ii)$ The amplitude of a point at $x = 0.375 \, m$ is:
$A = 0.06 \sin \left(\frac{2\pi}{3} \times 0.375\right)$
$A = 0.06 \sin \left(\frac{2\pi}{3} \times \frac{3}{8}\right)$
$A = 0.06 \sin \left(\frac{\pi}{4}\right)$
$A = 0.06 \times \frac{1}{\sqrt{2}} \approx 0.0424 \, m$.

Explore More

Similar Questions

$A$ wave is reflected from a rigid support. The change in phase on reflection will be

$A$ standing wave having $3$ nodes and $2$ antinodes is formed between two atoms having a distance of $1.21 \; \mathring{A}$ between them. The wavelength of the standing wave is .... $\mathring{A}$

Consider the three waves $z_1, z_2$ and $z_3$ as $z_1 = A \sin(kx - \omega t)$,$z_2 = A \sin(kx + \omega t)$ and $z_3 = A \sin(ky - \omega t)$. Which of the following represents a standing wave?

$A$ sound source of frequency $170 \, Hz$ is placed near a wall. $A$ man walking from the source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is $340 \, m/s$,the distance (in metres) separating the two adjacent positions of minimum intensity is

Stationary waves of frequency $300\, Hz$ are formed in a medium in which the velocity of sound is $1200\, m/s$. The distance between a node and the neighbouring antinode is ... $m$.

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