Question

a longitudinal progressive wave is given by the equation y=5×10^-2sinπ(400t+x)m.find 1.amplitude 2.frequncy 3.wave length 4.velocity of the wave 5.velocity and acceleration of the particle at x=1/6m at t=0.01s 6.maximum particle velocity and acceleration.​

Answer 1

Answer:

y=Asin(ωt+kx)

y=5×10^−2sin(πt×400+πx)

Comparing, we get A=5×10^−2 and k=π

λ=2π/k

λ=2 m

Ans=5×10−2m, λ=2m

hope it's help you..

Answer 2

$\rm y = 5 \times {10}^{ - 2} \sin\pi(400t + x)$

$\rm \implies y = 5 \times {10}^{ - 2} \sin(400\pi t + \pi x)$

Now, amplitude is max displacement :

$\boxed{ \bf amplitude = 5 \times {10}^{ - 2} \: m}$

Now, let frequency be f :

$\rm \: f = \dfrac{1}{T}$

$\rm \implies \: f = \dfrac{ \omega}{2\pi}$

$\rm \implies \: f = \dfrac{400\pi}{2\pi}$

$\boxed{ \bf\implies \: f = 200 \: hz}$

Now, let wavelength be $\lambda$.

$\rm \lambda = \dfrac{2\pi}{k}$

$\rm \implies \lambda = \dfrac{2\pi}{\pi}$

$\boxed{ \bf \implies \lambda = 2 \: m}$

Let velocity be v :

$\rm y = 5 \times {10}^{ - 2} \sin(400\pi t + \pi x)$

$\rm \implies v = 20\pi\cos(400\pi t + \pi x)$

At x = 1/6 m and t = 0.01 sec:

$\rm \implies v = 20\pi\cos(4\pi + \frac{\pi}{6} )$

$\rm \implies v = 20\pi\cos( \dfrac{\pi}{6} )$

$\boxed{ \bf \implies v = 54.38 \: m {s}^{ - 1} }$

Hope It Helps.