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Equations of a stationary wave and a travelling wave are $y_{1}=9 \sin k x coswt$ and $y_{2}=9 \sin (\omega t-k x)$ The Phase difference between two points $x_{1}=\frac{4 \pi}{7 k}$ and $x_{2}=\frac{3 \pi}{5 k}$ is $\varphi_{1}$ for the first wave and $\varphi_{2}$ for the second wave. The ratio $\varphi_{1} / \varphi_{2}$
Option: 1
$\frac{35}{1}$

Option: 2
$\frac{30}{1}$

Option: 3
$\frac{1}{35}$

Option: 4
$\frac{35}{2}$

Answers (1)

best_answer

$y=a \sin k x \text { cos wt }$

$\varphi t=\pi$

$x_{1}=\frac{4 \pi}{7 k} \quad x_{2}=\frac{3 \pi}{5 K}$

$y_{2}=a \sin (\omega t-k x)$

$\varphi_{2}=K\left(x_{2}-x_{1}\right)$

$\varphi_{2}=k\left[\frac{3 \pi}{5 k}-\frac{4 \pi}{7 k}\right]$

$\begin{aligned} & \varphi_{2}=\left(\frac{21-20}{35}\right)^{\pi} \\ & \varphi_{2}=\frac{\pi}{35} \\ & \frac{\varphi_{1}}{\varphi_{2}}=\frac{\pi}{\frac{\pi}{35}}=\frac{35 \pi}{\pi}=\frac{35}{1} \\ & {\left[\frac{\varphi_{1}}{\varphi_{2}}=\frac{35}{1}\right]} \end{aligned}$

Posted by

Rakesh

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