Question 536: Waves

Question

A transverse harmonic disturbance is 
produced in a string. The maximum 
transverse velocity is 3 m/s and the maximum 
transverse acceleration is 90 m/s$$^2$$
. If the wave velocity is 20 m/s, then find the waveform.

Accepted Answer

Step 1: Understanding Symbols
Let $$\omega$$ be the angular velocity and $$A$$ be the amplitude.
Step 2: Maximum Velocity
The highest speed up and down (maximum transverse velocity) is $$V_{\max} = \omega A = 3~\mathrm{m/s}$$.
Step 3: Maximum Acceleration
The highest acceleration up and down is $$\omega^2 A = 90~\mathrm{m/s}^2$$.
Step 4: Wave Velocity Relation
The speed that the wave itself travels is $$\frac{\omega}{k} = 20~\mathrm{m/s}$$, where $$k$$ is called the wave number.
Step 5: Find Angular Velocity $$\omega$$
To find $$\omega$$, divide the maximum acceleration equation by the maximum velocity equation:
$$\frac{A \omega^2}{A \omega} = \frac{90}{3}$$
So, $$\omega = 30~\mathrm{rad/s}$$.
Step 6: Find Amplitude $$A$$
From the maximum velocity equation: $$A = \frac{3}{30} = 0.1~\mathrm{m}$$.
Step 7: Find Wave Number $$k$$
We know $$k = \frac{\omega}{v}$$, where $$v$$ is the wave speed.
$$k = \frac{30}{20} = 1.5~\mathrm{m}^{-1}$$
Step 8: Write the Wave Equation
The wave can be written as: $$y = A \sin(\omega t + k x + \phi)$$, where $$\phi$$ is a starting phase (it can take any value).
So, the waveform is: $$y = 0.1 \sin(30t + 1.5x \pm \phi)$$

Answer

$$[y=(0.45) \sin \{[(30) t \pm(1.5) x] \pm \phi\}]$$

Answer

$$[y=(0.5) \sin \{[(50) t \pm(2.5) x] \pm \phi\}]$$

Answer

$$[y=(0.1) \sin \{[(30) t \pm(1.5) x] \pm \phi\}]$$

Answer

$$[y=(0.2) \sin \{[(60) t \pm(1.5) x] \pm \phi\}]$$

Waves

A transverse harmonic disturbance is produced in a string. The maximum transverse velocity is 3 m/s and the maximum transverse acceleration is 90 m/s$^2$ . If the wave velocity is 20 m/s, then find the waveform.

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