Question

The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive x-direction is given by f(t) = A sin$\left(\frac{t}{T}\right)$ The wave speed is v. Write the wave equation.

• A. $y=Asin(\frac{t}{T}-\frac{x}{vT})$
• B. $y=Asin(\frac{2\pi t}{T}-\frac{2\pi x}{vT})$
• C. $y=Asin\left(\frac{x}{vT}\right)$
• D. None of these

Answer 1

The correct option is A

$y=Asin(\frac{t}{T}-\frac{x}{vT})$

We are given f(t) = A sin $\left(\frac{t}{T}\right)$

Which implies the amplitude is A and comparing it with standard

SHM equation (y = A sin $\omega t$)

We get $\omega$= $\frac{1}{T}$

Now we know a sinusoidal wave moving in +ve x direction looks like y = A sin (kx - $\omega$t) or y = A sin ($\omega$t - kx) [Basically the signs of the two terms containing x and t respectively must be opposite]

Also$k=\frac{\omega }{v}=\frac{1}{vT}$

$\Rightarrow$for given wave equation is $y=Asin(\frac{t}{T}-\frac{x}{vT})$