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A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the - Physics

Sum

A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m s−1 and the amplitude is 0⋅5 cm. (a) Find the wavelength and the frequency. (b) Write the equation giving the displacement of different points as a function of time. Choose the X-axis along the string with the origin at one end and  t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.

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Solution

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm
(a) Wavelength and frequency of the string when it is vibrating in its 1st overtone (n = 2):
\[L = \frac{n\lambda}{2}\]

\[\Rightarrow \lambda = L = 2  m\] 

\[ \Rightarrow f = \frac{\nu}{\lambda} = \frac{200}{2} = 100  \text{ Hz  }\]
(b) The stationary wave equation is given by:

\[y = 2A  \cos\frac{2\pi x}{\lambda}\sin\frac{2\pi vt}{\lambda}\] 

\[ = 0 . 5\cos\frac{2\pi x}{2}\sin\frac{2\pi \times 200  t}{2}\] 

\[= \left( 0 . 5  cm \right)\cos\left[ \left( \pi m^{- 1} \right)  x \right]\sin\left[ \left( 200\pi s^{- 1} \right)  t \right]\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 51 | Page 327
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