A travelling harmonic wave on a string is described by y(x, t)=7.…

Question

A travelling harmonic wave on a string is described by
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
What are the displacement and velocity of oscillation of a point at x = 1cm, and t = 1s? Is this velocity equal to the velocity of wave propagation?

✓

Answer

The given harmonic wave is:
$\text{y}(\text{x, t})=7.5\sin\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
For x = 1cm and t = 1s,
$\text{y}=(1,1)=7.5\sin\Big[0.0050+12+\frac{\pi}{4}\Big]$
$=7.5\sin\Big[12.0050+\frac{\pi}{4}\Big]$
$=7.5\sin\theta$
Where, $\theta=12.0050+\frac{\pi}{4}=12.0050+\frac{3.14}{4}=12.79\text{ rad}$
$=\frac{180}{3.14\times12.79}=732.81^\circ$
$\therefore\ \text{y}=(1,1)=7.5\sin[732.81^\circ]$
$=7.5\sin(90\times8+12.81^\circ)$
$=7.5\sin(12.81^\circ)$
$=7.5\times0.2217$
$=1.6629\approx1.663\text {cm}$
The velocity of the oscillation at a given point and time is given as:
$\text{v}=\frac{\text{d}}{\text{dt}}\text{y}(\text{x, t})=\frac{\text{d}}{\text{dt}}\Big[7.5\sin\big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\big)\Big]$
$=7.5\times12\cos\Big(0.0050\text{x}+12\text{t}+\frac{\pi}{4}\Big)$
At x = 1cm and t = 1s:
$\text{v}=\text{y}(1, 1)=90\cos\Big(12.005+\frac{\pi}{4}\Big)$
$=90\cos(732.81^\circ)=90\cos(90\times8+12.81^\circ)$
$=90\cos(12.81^\circ)$
$=90\times0.975=87.75\text{cm/s}$
Now, the equation of a propagating wave is given by:
$\text{y}(\text{x, t})=\text{a}\sin(\text{kx}+\text{wt}+\phi)$
Where,
$\text{k}=\frac{2\pi}{\lambda}$
$\therefore\ \lambda=\frac{2\pi}{\text{k}}$
And $\omega=2\pi\text{ v}$
$\therefore\ \text{v}=\frac{\omega}{2\pi}$
Speed $=\text{v}=\text{v}\lambda=\frac{\omega}{\text{k}}$
Where
$\omega=12\text{ rad/s}$
$\text{k}=0.0050\text{m} ^{-1}$
$\therefore\ \text{v}=\frac{12}{0.0050}=2400\text{cm/s}$
$\therefore$ Hence, the velocity of the wave oscillation at x = 1cm and t = 1s is not equal to the velocity of the wave propagation.

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