Question

The mathematical forms for three sinusoidal travelling waves are given by
Wave $1:y(x,t)=\left(2cm\right)sin(3x-6t)$
Wave $2:y(x,t)=\left(3cm\right)sin(4x-12t)$
Wave $3:y(x,t)=\left(4cm\right)sin(5x-11t)$
where $x$ is in meters and $t$ is in seconds. Of these waves

• A. Wave $1$ has the greatest wave speed and the greatest maximum transverse string speed.
• B. Wave $2$ has the greatest wave speed and wave $1$ has the greatest maximum transverse string speed.
• C. Wave $3$ has the greatest wave speed and the greatest maximum transverse string speed.
• D. Wave $2$ has the greatest maximum transverse string speed.

Answer 1

The correct option is C Wave $3$ has the greatest wave speed and the greatest maximum transverse string speed.

Comparing the given wave equation with the standard form of wave equation $y=A\sin (kx-\omega t)$, we get

For wave 1: $A=2,k=3,\omega =6$

So, wave speed $v=\frac{\omega }{k}=6/3=2cm/s$

The magnitude of maximum transverse velocity, $u_{max}=\frac{dy}{dt}{|}_{max}=A\omega =2\left(6\right)=12cm/s$

For wave 2: $A=3,k=4,\omega =12$

So wave speed $v=\frac{\omega }{k}=12/4=3cm/s$

The magnitude of maximum transverse velocity, $u_{max}=\frac{dy}{dt}{|}_{max}=A\omega =3\left(12\right)=36cm/s$

For wave 3: $A=4,k=5,\omega =11$

So wave speed $v=\frac{\omega }{k}=11/5cm/s$

The magnitude of maximum transverse velocity, $u_{max}=\frac{dy}{dt}{|}_{max}=A\omega =4\left(11\right)=44cm/s$