# 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?

#### 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?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 8 | Page 324