Question

The displacement of a plane progressive wave in a medium, traveling towards the positive x-axis with velocity 4m/s at t=0 is given by \(y=3sin2\pi(-\frac{x}{3})\). Then the expression for the displacement at a later time t=4 sec will be

• A. \(y=3\,sin2\pi(-\frac{x-16}{3})\)
• B. \(y=3\,sin2\pi(\frac{-x-16}{3})\)
• C. \(y=3\,sin2\pi(\frac{-x+1}{3})\)
• D. \(y=3\,sin2\pi(\frac{-x-1}{3})\)

Answer 1

The Correct Option is A

Wave Equation Analysis 

The given wave equation is:

\[ y = 3\sin(2\pi(-3x)) \]

This represents a plane progressive wave at time \( t = 0 \), traveling in the positive x-direction with velocity \( v = 4 \, \text{m/s} \).

To determine the form of the wave at a later time \( t \), we use the general equation of a progressive wave:

\[ y = A \sin(kx - \omega t + \phi) \]

From the original equation, we identify:

• Amplitude \( A = 3 \)
• Wave number \( k = 2\pi \cdot 3 = 6\pi \)
• So, \( \lambda = \frac{2\pi}{k} = \frac{1}{3} \, \text{m} \)

 

The wave speed is given by: \[ v = \frac{\omega}{k} = 4 \quad \Rightarrow \quad \omega = 4k = 4 \cdot 6\pi = 24\pi \]

So, the time-dependent wave equation becomes: \[ y = 3\sin(2\pi(-3x) - 24\pi t) \]

This can be simplified using the identity: \[ y = 3\sin\left[2\pi\left(-3x - 12t\right)\right] \]

Now, substituting \( t = \frac{4}{3} \, \text{s} \), we get: \[ y = 3\sin\left[2\pi\left(-3x - 12 \cdot \frac{4}{3}\right)\right] = 3\sin\left[2\pi\left(-3x - 16\right)\right] \]

This confirms that the wave moves correctly with time, and thus:

\[ y = 3\sin\left(2\pi\left(-\frac{x - 16}{3}\right)\right) \]

Therefore, the correct option is (A): \[ y = 3\sin\left(2\pi\left(-\frac{x - 16}{3}\right)\right) \]

Answer 2

General Wave Equation: 

The wave propagating in one dimension is given by:

\[ y = A\sin(kx - \omega t + \phi) \]

Where:

• \(A\) is the amplitude
• \(k\) is the wave number
• \(\omega\) is the angular frequency
• \(x\) is the position
• \(t\) is the time
• \(\phi\) is the phase constant

 

Given wave equation at \(t = 0\):

\[ y = 3\sin\left(2\pi\left(-\frac{x}{3}\right)\right) = 3\sin\left(-\frac{2\pi x}{3}\right) \]

This matches the form \(y = A\sin(\omega t - kx)\) with \(\omega t = 0\), \(A = 3\), and \(k = \frac{2\pi}{3}\). Since the equation has a negative sign with \(x\), the wave moves in the positive x-direction.

Wave speed:

\[ v = \frac{\omega}{k} = 4 \Rightarrow \omega = vk = 4 \cdot \frac{2\pi}{3} = \frac{8\pi}{3} \]

Wave equation at time \(t\):

\[ y = 3\sin\left(\frac{8\pi}{3}t - \frac{2\pi}{3}x\right) \]

At \(t = 4\) seconds:

\[ y = 3\sin\left(\frac{8\pi}{3} \cdot 4 - \frac{2\pi}{3}x\right) = 3\sin\left(\frac{32\pi}{3} - \frac{2\pi}{3}x\right) \]

Factoring out \(2\pi\):

\[ y = 3\sin\left[2\pi\left(\frac{-x + 16}{3}\right)\right] = 3\sin\left[2\pi\left(-\frac{x - 16}{3}\right)\right] \]

Therefore, the correct option is (A):

\[ y = 3\sin\left(2\pi\left(-\frac{x - 16}{3}\right)\right) \]