Question

A string of length $ 20 cm $and linear mass density $ 0.40 gc{m}^{-1}$ is fixed at both ends and is kept under a tension of $ 16 N$. A wave pulse is produced at $ t=0$ near an end as shown in figure (below), which travels towards the other end.

(a) When will the string have the shape shown in the figure again?

(b) Sketch the shape of the string at a time half of that found in part (a).

20em 

Answer 1

Step 1: Given data

Length of the string, $L=20cm=0.2m$

Linear mass density, $m=0.40gc{m}^{-1}$

$$\begin{gathered} =0.40\times {10}^{-3}/{10}^{-2} \\ =0.040kg/m \end{gathered}$$

Tension in the string, $T=16N$

Let Time is taken to reach other ends is$T,$ Time is taken to see the pulse again in the original position$T_1,$and the Time $T_2$ be a time half of that found in part (a)

Step 2: Formula used

Wave velocity,$v=\sqrt{\frac{T}{m}}$where $T$ is the tension and $m$ is the mass per unit length.

Step 3: (a) Find time taken by the string to gain shape again

Velocity of wave, $v=\sqrt{\frac{T}{m}}$

$$\begin{gathered} =\sqrt{\frac{16}{0.040}} \\ =20m/s \end{gathered}$$

Time is taken to reach other ends, $T=\frac{L}{v}$

$$\begin{gathered} =\frac{0.2}{20} \\ =0.01s \end{gathered}$$

Since it takes $T=0.01s$ to reach other ends hence to see the pulse again in the original position the pulse has to reach the end and again come back to the first position hence the time taken will be twice of$T=0.01s$,

Therefore, Time is taken to see the pulse again in the original position$T_1=2T$

$\Rightarrow T_1=2\times 0.01=0.02s$

Hence, the string will have the shape shown in the figure again after$0.02s$.

Step 4: (b) Sketch the shape of the string at a time half of that found in part (a)

The shape of the string at a time half of that found in part (a) That is,

$$\begin{gathered} T_2=\frac{T_1}{2} \\ \Rightarrow T_2=\frac{0.02}{2} \\ \Rightarrow T_2=0.01S \end{gathered}$$

Hence, at $T_2=0.01s$ there will be a trough at the right end as it is reflected.