Question

A wave traveling on a string at a speed of $10m{s}^{-1}$ causes each particle of the string to oscillate with a time period of $20ms$.

(a) What is the wavelength of the wave?

(b) If the displacement of a particle is $15mm$ at a certain instant, what will be the displacement of a particle $10cm$ away from it at the same instant?

Answer 1

Explanation for the correct answer is,

Step 1. Given Data:

$$\begin{gathered} v=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \\ \Rightarrow T=20\times {10}^{-3}sec \\ \Rightarrow \lambda =Wavelengthofwave \\ \Rightarrow \varphi =Phasedifference \end{gathered}$$

Step 2. (a) Wavelength is obtained as:

$$\begin{gathered} v=\lambda /T \\ \Rightarrow \lambda =vt \\ \Rightarrow \lambda =20cm \end{gathered}$$

Step 3. (b) Phase shift difference:

$$\begin{gathered} \varphi =\left(\frac{2\pi }{\lambda }\right)x \\ \Rightarrow \varphi =\frac{2\pi }{20}x10 \\ \Rightarrow \varphi =\pi rad \end{gathered}$$

Step 4. Further:

$$\begin{gathered} y_1=a\sin (\omega t–kx) \\ \Rightarrow 1.5=a\sin (\omega t–kx) \end{gathered}$$

The displacement of a particle at$x=10cm$

$$\begin{gathered} \Rightarrow a\sin (\omega t–kx+\pi )=y_2 \\ \Rightarrow -a\sin (\omega t–kx)=-1.5mm \end{gathered}$$

Hence, (a) $\lambda =20cm$ & (b) $y_2=-1.5mm$.