Chapter 11, Problem 22P

(a)

To determine

To Find: The amplitude.

Answer to Problem 22P

The amplitude of A wave is $2.5\text{ m}$ and the amplitude of B wave is $3.5\text{ m}$

Explanation of Solution

Given:

Graph is given in the equation.

Formula used:

A wave and B wave intersects at y axis, the point of intersection will give us the amplitude of A wave and B wave respectively.

Calculation:

By identifying these two wave, amplitude of A wave is $2.5\text{ m}$ and amplitude of B wave is $3.5\text{ m}$ is directly obtained from the graph.

Conclusion:

Thus, the amplitude of A wave is $2.5\text{ m}$ and the amplitude of B wave is $3.5\text{ m}$ .

(b)

To determine

To Find: The frequency.

Answer to Problem 22P

The frequency of A wave is $f_a=0.25\text{ Hz}$ and frequency of B wave is $f_b=0.5\text{ Hz}$ .

Explanation of Solution

Given:

Completion time of 1 cycle for both the wave is easily identified from the graph.

A wave takes 4 seconds to complete 1 cycle (T_A= 4 sec).

B wave takes 2 seconds to complete 1 cycle (T_B= 2 sec).

Formula used:

The relationship between the period and frequency is,

  $\text{Frequency }f\text{ = }\frac{\text{1}}{\text{period }(T)}$

Calculation:

From the above equation,

Frequency of A wave $f_A=\frac{1}{4}=0.25\text{ Hz}$

Frequency of B wave $f_B=\frac{1}{2}=0.5\text{ Hz}$

Conclusion:

Thus, the frequency of A wave is $f_a=0.25\text{ Hz}$ and frequency of B wave is $f_b=0.5\text{ Hz}$ .

(c)

To determine

To Find: The period.

Answer to Problem 22P

The period of A wave is $4\sec$ and period of B wave is $2\sec$ .

Explanation of Solution

Given:

Graph is given for A and B.

Formula used:

The period of any wave is the amount of time taken to complete one cycle.

Calculation:

From the graph, it is clearly visible that the period of wave A and wave B is 4 seconds and 2 seconds respectively.

  $\begin{array}{l}{T}_A=4\text{ sec}\\ T_B=2\text{ sec}\end{array}$

Conclusion:

Thus, the period of A wave is $4\text{ sec}$ and period of B wave is $2\text{ sec}$ .

(d)

To determine

To Derive: The equation in the form of sine and cosine.

Answer to Problem 22P

The equation for the wave A is $X_A=2.5\sin \left(\frac{\pi }{2}\right)t$ and the equation for the wave B is $X_B=3.5\cos \left(\pi t\right)$

Explanation of Solution

Given:

Amplitude of wave $A_A=2.5\text{ m}$

Amplitude of wave $A_B=3.5\text{ m}$

Frequency of wave A $f_A=\frac{1}{4}=0.25\text{ Hz}$

Frequency of wave B $f_B=\frac{1}{2}=0.5\text{ Hz}$

Formula used:

Displacement is a function of time, which is represented by following two way.

  $\begin{array}{l}X=A\cos (\omega t)\\ X=A\sin (\omega t)\end{array}$

Calculation:

Wave A has zero displacement at time t = 0, so it is having sine function and it is as follows.

  $\begin{array}{l}{X}_A=A_A\sin (\omega t)\\ X_A=A_A\sin (2\times \pi \times f_{A}t)\\ X_A=2.5\sin \left(\frac{\pi }{2}\right)t\end{array}$

Wave B has maximum displacement at time t = 0, so it is having cosine function and it is as follows.

  $\begin{array}{l}{X}_B=A_B\cos (\omega t)\\ X_B=A_B\cos (2\times \pi \times f_{B}t)\\ X_B=3.5\cos \left(\pi \right)t\end{array}$

Conclusion:

Thus, the equation for the wave A is $X_A=2.5\sin \left(\frac{\pi }{2}\right)t$ and the equation for the wave B is $X_B=3.5\cos \left(\pi t\right)$