Question

A string of linear mass density 0⋅5 g cm⁻¹ and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure 15-E4). The ring can move on a vertical rod. A wave pulse is produced on the string which moves towards the ring at a speed of 20 cm s⁻¹. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to region its shape. (b) The shape of the string changes periodically with time. Find this time period. (c) What is the tension in the string?

Answer 1

Given,
Linear mass density of the string = 0.5 gcm⁻¹
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms⁻¹



The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.
$\mathrm{Phase}\mathrm{change}=0$
(a) $$\begin{gathered} \mathrm{Total}\mathrm{distance},S=20+20=40\mathrm{cm} \\ \mathrm{Wave}\mathrm{speed},\nu =20\mathrm{m}/\mathrm{s} \end{gathered}$$
Time taken to regain shape:
$\mathrm{Time}=\frac{S}{\nu }=\frac{40}{20}=2\mathrm{s}$

(b) The wave regain its shape after covering a period distance$=2\times 30=60$ cm
$\therefore \mathrm{Time}\mathrm{period}=\frac{60}{20}=3\mathrm{s}$

(c) Frequency, $n=\frac{1}{\mathrm{Time}\mathrm{period}}=\frac{1}{3}{\mathrm{s}}^{-1}$
We know:
$n=\frac{1}{2l}\sqrt{\left(\frac{T}{m}\right)}$
Here, T is the tension in the string.
Now,

$$\begin{gathered} m=\mathrm{Mass}\mathrm{per}\mathrm{unit}\mathrm{length} \\ =0.5\mathrm{gm}/\mathrm{cm} \\ \Rightarrow \frac{1}{3}=\frac{1}{\left(2\times 30\right)}\sqrt{\left(\frac{T}{0.5}\right)} \\ \Rightarrow T=400\times 0.5 \\ =200\mathrm{dyn} \\ =2\times {10}^{-3}\mathrm{N} \end{gathered}$$