Question

A travelling wave y = 5 sin$\left( 20\text{ t}-50\text{ x} \right)$ is moving along the x axis in a taut string (here x, y, t are in S.I. units). The speed of propagation in $\text{cm/sec}$ is .

Answer 1

The given equation of travelling wave is
Y = 5 sin$\left( 20\text{ t}-50\text{ x} \right)$
When we compare it with the standard equation, the value of w and k can be calculated.
Now, as $\text{v}=\text{v}\lambda \text{=}\dfrac{\lambda }{\text{T}}$ so by putting values of $\lambda =\left[ \dfrac{2\pi }{\text{R}} \right]$ and $\text{T}=\left[ \dfrac{2\pi }{\text{w}} \right]$ , we get value of v.

Complete step by step solution:
Given equation is y = 5 sin$\left( 20\text{ t}-50\text{ x} \right)$
Comparing it with standard equation:
Y = r sin$\left( \text{w t}-\text{k x} \right)$
We get:
r = 5$\text{m}$
w = 20
k = 50
Now, we know that
$\text{w}=\dfrac{2\pi }{\text{T}}$ Or $\text{T}=\dfrac{2\pi }{\text{w}}$ …….. (1)
And
$\text{k=}\dfrac{\text{2 }\\!\\!\pi\\!\\!\text{ }}{\text{ }\\!\\!\lambda\\!\\!\text{ }}$ Or $\text{ }\\!\\!\lambda\\!\\!\text{ =}\dfrac{2\pi }{\text{R}}$ …… (2)
Also, v = V $\text{ }\\!\\!\lambda\\!\\!\text{ }$
Putting values of $\text{ }\\!\\!\lambda\\!\\!\text{ }$and T from equations (1) and (2), we get:
$\begin{aligned} & \text{v}=\left( \dfrac{2\pi }{\text{k}} \right)\left( \dfrac{\text{w}}{2\pi } \right) \\\ & \Rightarrow \dfrac{\text{w}}{\text{k}} \\\ \end{aligned}$
$\Rightarrow \dfrac{20}{50}$
$\Rightarrow \text{v}=\dfrac{2}{5}$
$\Rightarrow 0\cdot 4\text{ m}$
$\therefore \text{v}=\text{40 cm}$

Note: Travelling waves are observed when the wave is not confined to a given space along the medium. The most commonly observed travelling wave is an ocean wave. A travelling wave is described by the equation $\text{y}\left( \text{x, t} \right)=\text{r sin}\left( \text{w t}-\text{k x} \right)$
Where
r= amplitude of the wave
w = angular frequency $=\dfrac{2\pi }{\text{T}}$
k = wave number $\text{=}\dfrac{\text{2}\\!\\!\pi\\!\\!\text{ }}{\text{ }\\!\\!\lambda\\!\\!\text{ }}$