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Wave Motion and Waves on a String — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsWave Motion and Waves on a String5 Marks
Question
A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is 1.0cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10kg/m and it is kept under a tension of 90N.
  1. Find the speed and the wavelength of the wave.
  2. 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.
  3. Find the velocity and acceleration of the particle at x - 50cm at time t = 10ms.
✓

Answer

Amplitude, $\text{A}=1\text{cm},$ Tension $\text{T}=90\text{N}$
Frequency, $\text{f}=\frac{200}{2}=100\text{Hz}$
Mass per unit length, $\text{m}=0.1\text{kg/mt}$
  1. $\Rightarrow\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}=30\text{m/s}$
$\lambda=\frac{\text{v}}{\text{f}}=\frac{30}{100}=0.3\text{m}=30\text{cm}$
  1. The wave equation $\text{y}=(1\text{cm})\cos2\pi\Big(\frac{\text{t}}{0.01\text{s}}\Big)-\Big(\frac{\text{x}}{30}\text{cm}\Big)$
[because at x = 0, displacement is maximum]
  1. $\text{y}=1\cos2\pi\Big(\frac{\text{x}}{30}-\frac{\text{t}}{0.01}\Big)$
$\Rightarrow\text{v}=\frac{\text{dy}}{\text{dt}}=\Big(\frac{1}{0.01}\Big)\sin2\pi\Big\{\big(\frac{\text{x}}{30}\big)-\big(\frac{\text{t}}{0.01}\big)\Big\}$

$\text{a}=\frac{\text{dv}}{\text{dt}}=-\Big\{\frac{4\pi^2}{(0.01)^2}\Big\}\cos2\pi\Big\{\big(\frac{\text{x}}{30}\big)-\big(\frac{\text{t}}{0.01}\big)\Big\}$

When, $\text{x}=50\text{cm},\ \text{t}=10\text{ms}=10\times10^{-3}\text{s}$

$\text{x}=\Big(\frac{2\pi}{0.01}\Big)\sin2\pi\Big\{\big(\frac{5}{3}\big)-\big(\frac{0.01}{0.01}\big)\Big\}$

$=\Big(\frac{\text{p}}{0.01}\Big)\sin\bigg(\frac{2\pi}{\frac{2}{3}}\bigg)=\Big(\frac{1}{0.01}\Big)\sin\Big(\frac{4\pi}{3}\Big)\\=-200\pi\sin\Big(\frac{\pi}{3}\Big)=-200\pi\text{x}\Big(\frac{\sqrt{3}}{2}\Big)$

$=544\text{cm/s}=5.4\text{m/s}$

Similarly

$\text{a}=\Big\{\frac{4\pi^2}{(0.01)^2}\Big\}\cos2\pi\Big\{\big(\frac{5}{3}\big)-1\Big\}$

$=4\pi^2\times10^4\times\frac{1}{2}$

$\Rightarrow2\times10^5\text{cm/s}^2$

$\Rightarrow2\text{km/s}^2$

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