Question

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1⋅0 and the displacement becomes zero 200 times per second. The linear mass density of the string is 0⋅10 kg m⁻¹ and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10 ms.

Answer 1

Given,
Amplitude of the wave = 1 cm
Frequency of the wave,
\[f = \frac{200}{2} = 100  \text{ Hz }\]
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N
(a) Velocity of the wave is given by
\[v = \sqrt{\frac{T}{m}}\]
Thus, we have:\[v = \sqrt{\left( \frac{90}{0 . 1} \right)} = 30  m/s\]
Now,
\[\text{ Wavelength, }   \lambda = \frac{v}{f} = \frac{30}{100} = 0 . 3  m\]
\[ \Rightarrow \lambda = 30  cm\]
(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
\[y = \left( 1  cm \right)\cos2\pi\left\{ \left( \frac{t}{0 . 01  s} \right) - \left( \frac{x}{30  cm} \right) \right\}\]  ...(1)
(c) Using \[\cos\left( - \theta \right) = \cos\theta\] 
in equation (1), we get: 
\[y = 1\cos2\pi\left( \frac{x}{30} - \frac{t}{0 . 01} \right)\]
\[Velocity,   v = \frac{dy}{dt}\]
\[ \Rightarrow v = \left( \frac{2\pi}{0 . 01} \right)\sin2\pi\left\{ \frac{x}{30} - \frac{t}{0 . 01} \right\}\]
And,
Acceleration,   \[a = \frac{d\nu}{dt}\]
\[ \Rightarrow a = \left\{ \frac{4 \pi^2}{\left( 0 . 01 \right)^2} \right\}\cos2\pi\left\{ \left( \frac{x}{30} \right) - \left( \frac{t}{0 . 01} \right) \right\}\]
\[\text{ When  x = 50  cm,   t = 10  ms = 10 \times  {10}^{- 3}   s .}\]
Now,
\[v = \left( \frac{2\pi}{0 . 01} \right)\sin2\pi\left\{ \left( \frac{5}{3} \right) - \left( \frac{0 . 01}{0 . 01} \right) \right\}\]
\[     = \left( \frac{2\pi}{0 . 01} \right)\sin\left( 2\pi \times \frac{2}{3} \right)\]
\[=  - \left( \frac{2\pi}{0 . 01} \right)\sin\frac{4\pi}{3}\]
\[=  -   200\pi\sin\frac{\pi}{3}\]
\[=  - 200\pi \times \frac{\sqrt{3}}{2}\]
\[=  - 544  cm/s\]
\[=  - 5 . 4  m/s\]
In magnitude, v = 5.4 m/s.
Similarly,
\[a = \left\{ \frac{4 \pi^2}{\left( 0 . 01 \right)^2} \right\}\cos2\pi\left\{ \left( \frac{5}{3} \right) - 1 \right\}\]
\[     = 4 \pi^2  \times  {10}^4  \times \frac{1}{2}\]
\[     \approx 2 \times  {10}^5   cm/ s^2   \text{ or  2  km}/ s^2\]