Answer to Question #85567 in Mechanics | Relativity for Rinku Rai

Question

The equation of transverse wave on a rope is
y(x, t) = 5sin (4.0t − 0.02x)
where y and x are measured in cm and t is expressed in second. Calculate the
maximum speed of a particle on the rope

Answer 1

The wave function for the transverse wave on the rope can be written as follows:

$$y(x,t) = A sin(\omega t - kx) = 5 sin(4.0 t - 0.02x) ,$$

here, $A = 0.05 m$ is the amplitude of the transverse wave, $\omega = 4.0 \dfrac{rad}{s}$ is the angular frequency of the transverse wave and $k = 0.02 \dfrac{rad}{cm} \cdot \dfrac{100 cm}{1 m} = 2 \dfrac{rad}{m}$ is the wavenumber.

By the definition, the velocity is the derivative of position with respect to time:

$$v = \dfrac{dy}{dt}.$$

So, we can write the speed of a particle on the rope as follows:

$$v(x, t) = \dfrac{dy(x,t)}{dt} = \dfrac{d}{dt}(A sin(\omega t - kx)) = A \omega cos(\omega t - kx).$$

As we can see from the formula, the speed of a particle on the rope is maximum when $cos(\omega t - kx) = \pm 1$.

Therefore,

$$v_{max} = A \omega = 0.05 m \cdot 4.0 \dfrac{rad}{s} = 0.2 \dfrac{m}{s}.$$

Answer:

$v_{max} = 0.2 \dfrac{m}{s}.$