Question 135: Waves

Question

The equation of a transverse wave travelling along a string is $y(x, t)=4.0 \sin \left[20 	imes 10^{-3} x+600 tight] \mathrm{mm}$, where $x$ is in mm and $t$ is in second. The velocity of the wave is :

Accepted Answer

Let's analyze the wave equation step by step.
The given wave is:
$$y(x, t) = 4.0 \sin \left[20 	imes 10^{-3} x + 600 tight] 	ext{ mm}.$$
First, simplify the coefficient of $ x $:
$$20 	imes 10^{-3} = 0.02 \, 	ext{mm}^{-1}.$$
So the equation becomes:
$$y(x, t) = 4.0 \sin\left(0.02x + 600tight) 	ext{ mm}.$$
A standard form for a travelling wave is:
$$y(x, t) = A \sin(kx - \omega t)$$
which represents a wave moving in the positive $ x $-direction with speed $ v = \frac{\omega}{k} $.
Notice that our wave equation has the form:
$$\sin(0.02x + 600t)$$
The positive sign in front of $ 600t $ means we can rewrite the phase as:
$$0.02x + 600t = 0.02x - (-600t),$$
which indicates that the angular frequency $ \omega $ in the standard form is effectively $ -600 $.
The velocity $ v $ of a wave is determined from the phase (for a constant phase, $ \phi = $ constant):
$$k x + \omega t = 	ext{constant}.$$
Differentiating with respect to $ t $:
$$k \frac{dx}{dt} + \omega = 0,$$
which gives:
$$\frac{dx}{dt} = -\frac{\omega}{k}.$$
Substituting the values:
$ k = 0.02 \, 	ext{mm}^{-1} $
$ \omega = 600 \, 	ext{s}^{-1} $
We have:
$$v = -\frac{600}{0.02} = -30000 	ext{ mm/s}.$$
Convert the velocity from mm/s to m/s:
$$-30000 \, 	ext{mm/s} = -30 \, 	ext{m/s}.$$
Thus, the velocity of the wave is $-30 \, 	ext{m/s}$.
The correct answer is Option D.

Answer

$-60 \mathrm{~m} / \mathrm{s}$

Answer

$+60 \mathrm{~m} / \mathrm{s}$

Answer

$+30 \mathrm{~m} / \mathrm{s}$

Answer

$-30 \mathrm{~m} / \mathrm{s}$

Waves

The equation of a transverse wave travelling along a string is y(x, t)=4.0 \sin \left[20 \times 10^{-3} x+600 t\right] \mathrm{mm}, where x is in mm and t is in second. The velocity of the wave is :

Choose the correct answer: