Velocity of Point on Transverse Sinusoidal Wave

To find the velocity of point $P$ on the string, we first determine the frequency, angular frequency, and magnitude of particle velocity, followed by its direction using the wave profile.

1. Calculation of Wave Parameters

Given:

• Wave speed, $v = 10\text{ cm/s} = 0.1\text{ m/s}$
• Wavelength, $\lambda = 0.5\text{ m}$
• Amplitude, $A = 10\text{ cm} = 0.1\text{ m}$
• Displacement at point $P$, $y = 5\text{ cm} = 0.05\text{ m}$

The angular frequency $\omega$ of the wave is given by:
$\omega = 2\pi f = 2\pi \left(\frac{v}{\lambda}\right) = 2\pi \left(\frac{0.1\text{ m/s}}{0.5\text{ m}}\right) = \frac{2\pi}{5}\text{ rad/s}$

2. Magnitude of Particle Velocity

The speed of a particle performing Simple Harmonic Motion at displacement $y$ is given by:
$|v_p| = \omega \sqrt{A^2 - y^2}$

Substituting the given values:
$|v_p| = \left(\frac{2\pi}{5}\right) \sqrt{(0.1)^2 - (0.05)^2} = \left(\frac{2\pi}{5}\right) \sqrt{0.0075} = \left(\frac{2\pi}{5}\right) \left(\frac{\sqrt{3}}{20}\right) = \frac{\sqrt{3}\pi}{50}\text{ m/s}$

3. Direction of Velocity

The particle velocity $v_y$ at any point on a transverse wave traveling in the positive $x$-direction is related to the slope of the wave profile by:
$v_y = -v \left(\frac{\partial y}{\partial x}\right)$

From the given snapshot of the wave:

• At point $P$, the string has a negative slope, i.e., $\frac{\partial y}{\partial x} < 0$.
• Since $v > 0$ (wave travels along $+x$), the particle velocity is:
  $v_y = - (+v)(-\text{slope}) > 0$

Thus, the particle at $P$ is moving upwards in the positive $y$-direction ($+\hat{j}$). Alternatively, as the wave shifts to the right, the crest approaches point $P$, causing its displacement to increase upwards.

Conclusion

The velocity vector of point $P$ is:
$\vec{v}_P = \frac{\sqrt{3}\pi}{50}\hat{j}\text{ m/s}$

Correct Answer: A