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Wave Motion and Waves on a String — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWave Motion and Waves on a String3 Marks
Question
The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by$\text{y}=(0.4\text{cm})\sin\big[(0.314\text{cm}^{-1})\text{x}\big]\cos\big[(600\pi\text{s}^{-1})\text{t}\big]$
  1. What is the frequency of vibration?
  2. What are the positions of the nodes?
  3. What is the length of the string?
  4. What is the wavelength and the speed of two travelling waves that can interfere to give this vibration ?
✓

Answer

The stationary wave equation is given by$\text{y}(0.4\text{cm})\sin\big[(0.314\text{cm}^{-1})\text{x}\big]\cos\big[(600\pi\text{s}^{-1})\text{t}\big]$
  1. $\omega=600\pi\Rightarrow2\pi\text{f}=600\pi\Rightarrow\text{f}=300\text{Hz}$
Wavelength,, $\lambda=\frac{2\pi}{0.314}=\frac{(2\times3.14)}{0.314}=20\text{cm}$
  1. Therefore nodes are located at, $0,\ 10\text{cm},\ 20\text{cm},\ 30\text{cm}$
  2. Length of the string $=\frac{3\lambda}{2}=3\times\frac{20}{2}=30\text{cm}$
  3. $\text{y}=0.4\sin(0.314\text{x})\cos(600\pi\text{t})$
$\Rightarrow0.4\sin\Big\{\big(\frac{\pi}{10}\big)\text{x}\Big\}\cos(600\pi\text{t})$
since $\lambda$ and $v$ are the wavelength and velocity of the waves that interfere to give this vibration $\lambda=20\text{cm}$
$\text{v}=\frac{\omega}{\text{k}}=6000\text{cm/sec}=60\text{m/s}$

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