Question

The displacement of a standing wave on a string is given by $y(x,t)=0.8\sin \left(x\right)\cos \left(20t\right)$ where $x$ and $y$ are in centimeters and $t$ is in seconds. Then:

• A. Frequency of the component waves is $10\text{Hz}$
• B. Wave speed of wave is $20\text{cm/s}$
• C. Amplitude of the component wave is $0.4\text{cm}$
• D. Particle velocity at $x=\frac{7\pi }{6}\text{cm}$ and at $t=\frac{\pi }{40}\text{sec}$ is $8\text{cm/s}$

Answer 1

Answer: (B), (C), (D)

Particle velocity at $x=\frac{7\pi }{6}\text{cm}$ and at $t=\frac{\pi }{40}\text{sec}$ is $8\text{cm/s}$

Given that
$y(x,t)=0.8\sin \left(x\right)\cos \left(20t\right)$

Comparing this with,
$y(x,t)=2A\sin \left(kx\right)\cos \left(\omega t\right)$

we get, Angular frequency $\left(\omega \right)=20\text{rad/s}$
Angular wave number $\left(k\right)=1{\text{cm}}^{-1}$
Frequency $\left(f\right)=\frac{\omega }{2\pi }=\frac{20}{2\pi }=\frac{10}{\pi }\text{Hz}$
Amplitude of component waves $\left(A\right)=0.4\text{cm}$
So, Wave speed $\left(v\right)=\frac{\omega }{k}=\frac{20}{1}=20\text{cm/s}$
Particle velocity $\left(v_p\right)=\frac{\partial y}{\partial t}=\frac{\partial }{\partial t}[0.8\sin \left(x\right)\cos \left(20t\right)]$
$\left(v_p\right)=-16\sin \left(x\right)\sin \left(20t\right)$
For, $x=\frac{7\pi }{6}\text{cm}$ and $t=\frac{\pi }{40}\text{sec}$
So, $v_p=-16\sin \left(\frac{7\pi }{6}\right)\sin (20\times \frac{\pi }{40})$
$v_p=-16\times \left(\frac{-1}{2}\right)=8\text{cm/s}$

Thus, options (b) , (c) and (d) are the correct answers.