Question

A harmonic wave on a rope is described by
y(x,t) = (4.0mm) sin [2π/0.82m ((10ms-1 ) 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 = 41.0 s

Answer 1

Answer on Question #72985, Physics / Mechanics | Relativity

Question. A harmonic wave on a rope is described by $y(x, t) = (4.0 \, \text{mm}) \sin \left[ \frac{2\pi}{0.82 \, \text{m}} \left( \frac{(10 \, \text{m/s}) \, t + x}{t} \right) \right]$.

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

ii) Determine the displacement and acceleration of the element of the rope located at $x = 0.58 \, \text{m}$ at time, $t = 41.0 \, \text{s}$.

Given. $x = 0.58 \, \text{m}; t = 41.0 \, \text{s}$.

Find. $\lambda, T, y(0.58, 41), a(0.58, 41) - ?$

Solution.

In accordance with the equation of wave

$$\xi = A \sin (\omega t + k x),$$

where

$$k = \frac{2\pi}{\lambda} = \frac{\omega}{v}$$

we have

$$\xi = A \sin (\omega t + k x) = A \sin k \left( \frac{\omega t}{k} + x \right) = A \sin \frac{2\pi}{\lambda} (v t + x).$$ $$\xi = y (x, t) = (4.0 \, \text{mm}) \sin \left[ \frac{2\pi}{0.82 \, \text{m}} \left( \left(10 \, \frac{\text{m}}{\text{s}}\right) t + x \right) \right]$$ $$\lambda = 0.82 \, \text{m}.$$ $$T = \frac{\lambda}{v} = \frac{0.82}{10} = 0.082 \, \text{s} = 82 \, \text{ms}.$$ $$y (0.58, 41) = 4.0 \cdot 10^{-3} \sin \left[ \frac{2\pi}{0.82} (10 \cdot 41 + 0.58) \right] =$$ $$= -0.00386 \, \text{m} = -3.86 \, \text{mm}.$$ $$a (x, t) = \frac{d^2 y (x, t)}{d t^2} = -4.0 \cdot 10^{-3} \left( \frac{2\pi}{0.82} \cdot 10 \right)^2 \sin \left[ \frac{2\pi}{0.82} (10 \cdot t + x) \right].$$ $$a (0.58, 41) = -4.0 \cdot 10^{-3} \cdot \left( \frac{2\pi}{0.82} \cdot 10 \right)^2 \cdot \sin \left[ \frac{2\pi}{0.82} (10 \cdot 41 + 0.58) \right] \approx 22.7 \, \frac{\text{m}}{\text{s}^2}.$$

Answer. $\lambda = 0.82 \, \text{m}; T = 82 \, \text{ms}; y(0.58, 41) = -3.86 \, \text{mm}; a(0.58, 41) = 22.7 \, \frac{\text{m}}{\text{s}^2}$.

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