Question

A harmonic wave on a rope is described by
y(x,t)=(4.00 mm)sin[2pi/0.82m ((10m/s)t+x)]

i) Calculate the wavelength and time period of the wave. ii) Determine the
displacement and acceleration of the element of the rope located at x = 0.58 m at
time, t = 0.41 s.

Answer 1

Answer on Question #73476 - Physics / Mechanics | Relativity

A harmonic wave on a rope is described by

$$y(x, t) = (4.00 \text{ mm}) \sin \left( \frac{2\pi}{0.82 \text{ m}} \left( \left(10 \frac{\text{m}}{\text{s}}\right) t + x \right) \right)$$

i) Calculate the wavelength and time period of the wave.

ii) Determine the displacement and acceleration of the element of the rope located at $x = 0.58$ m at time, $t = 0.41$ s.

Solution:

The general form of the wave equation

$$y(x, t) = A \sin \left( \frac{2\pi}{\lambda} (v t + x) \right)$$

i) Where $\lambda$ is a wavelength, $v$ is a wave velocity.

Thus

Wavelength $\lambda = 0.82$ m

Time period $T = \frac{\lambda}{v} = \frac{0.82}{10} = 0.082$ s

ii) The displacement at the $x = 0.58$ m and $t = 0.41$ s

$$y(x, t) = 4.00 \sin \left( \frac{2\pi}{0.82} (10 \times 0.41 + 0.58) \right) = -3.857 \text{ mm}$$

The acceleration

$$a = y'' = -0.004 \times \left( \frac{2\pi}{0.82} \times 10 \right)^2 \sin \left( \frac{2\pi}{0.82} (10 \times 0.41 + 0.58) \right) = 22.65 \frac{\text{m}}{\text{s}^2}$$

Answers:

Wavelength $\lambda = 0.82$ m

Time period $T = \frac{\lambda}{v} = \frac{0.82}{10} = 0.082$ s

The displacement $y(x, t) = -3.857$ mm

The acceleration $a = 22.65 \frac{\text{m}}{\text{s}^2}$