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The Equation of a Wave Travelling on a String is (A) in Which Direction Does - Physics

Definition
Sum

The equation of a wave travelling on a string is \[y = \left( 0 \cdot 10  \text{ mm } \right)  \sin\left[ \left( 31 \cdot 4  m^{- 1} \right)x + \left( 314  s^{- 1} \right)t \right]\]
(a) In which direction does the wave travel? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string?

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Solution

Given,
Equation of the wave,
\[y = \left( 0 . 10  \text{ mm } \right)  \sin\left( 31 . 4  m^{- 1} \right)x + \left( 314  s^{- 1} \right)  t\]
The general equation is \[y = A\sin\left\{ \left( \frac{2\pi x}{\lambda} \right) + \omega t \right\}\] 
From the above equation, we can conclude:
(a) The wave is travelling in the negative x-direction.
(b) \[\frac{2\pi}{\lambda} = 31 . 4   m^{- 1}\] 

\[\Rightarrow \lambda = \frac{2\pi}{31 . 4} = 0 . 2  m =   20  cm\]
And,
\[\omega = 314   s^{- 1} \] 

\[ \Rightarrow 2\pi f = 314\] 

\[ \Rightarrow f = \frac{314}{2\pi}\] 

\[= \frac{314}{2 \times 3 . 14}\] 

\[= 50   s^{- 1}  = 50  Hz\]
Wave speed:

\[\nu = \lambda f = 20 \times 50\] 

\[=1000  cm/s\]
(c) Maximum displacement, A = 0.10 mm

Maximum  velocity = \[a\omega = 0 . 1 \times  {10}^{- 1}  \times 314\] 

= 3.14  cm/s

  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 8 | Page 324
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