Question

A transverse sine wave moves along a string in the positive \( x \) direction at a speed of \( 20 cm / s \). The wavelength of the wave is \( 1 m \) and its amplitude is \( 10 cm \). The snapshot of the wave is shown in the figure. The acceleration and velocity of point \( P \) whose displacement at this instant is \( 5 cm \) are \( \overline{a_{P}} \) and \( \overline{V_{P}} \) respectively. Then \( \overline{V_{P}}=2 \sqrt{3} \pi \hat{j} cm / s \) \( \overline{V_{P}}=-2 \sqrt{3} \pi \hat{j} cm / s ^{2} \) \( \overrightarrow{a_{p}}=\frac{4 \pi^{2}}{5} \hat{j} cm / s ^{2} \) \( \overline{a_{p}}=-\frac{4 \pi^{2}}{5} \hat{j} cm / s ^{2} \)

Answer 1

Given λ = 1 m

A = 10 cm = .10 m

x = 5 cm

x = 0.05 m

v = 20 cm/s

v = .2 m/s

velocity of point P is

v_P = \(\omega\sqrt{A^2-x^2}\) 

v_P = \(\frac{2\pi v}{\lambda}\sqrt{A^2-x^2}\)

 v_P = \(\frac{2\times \pi\times0.20}1\sqrt{(.10)^2-(0.05)^2}\) 

 v_P = \(\frac{2\pi\times20}{100}\sqrt{(.10)^2-(0.05)^2}\)

 v_P = \(\frac{2\pi}{5}\sqrt{0.01-0.0025}\)

  v_P = \(\frac{2\pi}{5}\sqrt{0.0075}\)

  v_P = \(\frac{2\pi}{5}\times0.086\)

  v_P = \(\frac{0.54}{5}\)

v_P = 0.108 m/s

Since the direction of point p is toward the positive y axis, the velocity can be written as a water farm.

v_P = 0.108 j m/s

Accelaeration

a_P = \(\omega^2A\)

 a_P = \((\frac{2\pi v}\lambda)^2A\)

 a_P =\(\frac{4\pi^2 v^2}{\lambda^2}\times0.10\)

 a_P =\(4\pi^2\times\frac{2}{10}\times\frac{2}{10}\times\frac{10}{100}\)

a_P = \(\frac{4\pi^2}{250}\)

a_P = \(\frac{2}{125}\pi^2\) m/s²