Chapter 14, Problem 77A

(a)

To determine

The time it takes for the pulse to return back to center from the string end.

Answer to Problem 77A

  $2.4\times {10}^{-3}\text{s}$

Explanation of Solution

Given:

Speed of the wave: $265\text{m/s}$

Length of the string: $63\text{cm}$

Formula used:

To calculate the time for the pulse to move to the end of the string and then back to its center, then use:

  $t=\frac{d}{v}$

Where,

  $t$ : Time

  $d$ : Distance

  $v$ : Velocity

Calculation:

Convert length of string to from m to cm:

  $63\text{cm}=0.63\text{m}$

Substitute the values and solve:

  $t=\frac{d}{v}$

  $\begin{array}{l}t=\frac{\text{0}\text{.63}\text{m}}{\text{265}\text{m/s}}\\ t=0.0024\text{s}\end{array}$

  $t=2.4\times {10}^{-3}\text{s}$

Conclusion:

Hence,it takes $2.4\times {10}^{-3}\text{s}$ for the pulse to move for the end of the string and then return to its center.

(b)

To determine

The position of string when the pulses return to the original position.

Explanation of Solution

Introduction:

The pulses are reversed when it is reflected from a medium which is denseras compared to the initial position.

The pulses when travel back from a denser medium, gets reversed in the position and the string is below its resting location.

Conclusion:

Hence,the position of the pulses is reversed when it is reflected from a denser medium.

(c)

To determine

The position where the pulse will meet if one plucks the string of $15$ cmfrom one end of the string.

Answer to Problem 77A

It is $15$ cm from the other end and the distance travelled from the other end would be same.

Explanation of Solution

Introduction:

The distance travelled by the string from the other side will be same. In case one plucks any string from a particular length away from one end, the other end will also exist at the same distance.

In case when you pluck a string of particular length from one end, the other end will exist at the same distance.

Conclusion:

Hence, if one plucks a string, $15$ cm from one end of the string,it is $15$ cm from the other end and the distance that is travelled here are same.