Question

A sinusoidal wave with an amplitude of 2 cm is traveling in the negative x direction. The distance between two consecutive crests is 2 cm, and the angular frequency is $\pi s^{-1}$. What is the displacement of the particle located at x = 20.5 cm at t = 10 s? Assume that at t = 0 and x = 0 the particle was at its mean position moving downward. Both x and y are measured in centimeters.

• A. 0 cm
• B. 1 cm
• C. 2 cm
• D. 3 cm

Answer 1

Answer: (C)

2 cm

Answer 2

The general equation for a sinusoidal wave traveling in the negative x direction is $y(x, t) = A \sin(kx + \omega t + \phi)$.

Given amplitude $A = 2$ cm, wavelength $\lambda = 2$ cm, and angular frequency $\omega = \pi$ s⁻¹.

The wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{2} = \pi$ cm⁻¹.

So, the equation becomes $y(x, t) = 2 \sin(\pi x + \pi t + \phi)$.

Using the initial condition: At $t = 0$ and $x = 0$, the particle was at its mean position ($y=0$) moving downward ($\frac{\partial y}{\partial t} < 0$).

1. $y(0, 0) = 2 \sin(\phi) = 0 \implies \sin(\phi) = 0$. This means $\phi = 0$ or $\phi = \pi$.

2. The velocity is $\frac{\partial y}{\partial t} = A\omega \cos(kx + \omega t + \phi) = 2\pi \cos(\pi x + \pi t + \phi)$.

At $x = 0, t = 0$, $\left(\frac{\partial y}{\partial t}\right)_{(0,0)} = 2\pi \cos(\phi)$.

Since the particle is moving downward, $\left(\frac{\partial y}{\partial t}\right)_{(0,0)} < 0$, which implies $2\pi \cos(\phi) < 0$. As $2\pi > 0$, we need $\cos(\phi) < 0$.

Comparing with $\phi = 0$ or $\phi = \pi$:

• If $\phi = 0$, $\cos(0) = 1$ (not less than 0).

• If $\phi = \pi$, $\cos(\pi) = -1$ (less than 0).

Thus, the phase constant $\phi = \pi$.

Substitute $\phi = \pi$ into the wave equation: $y(x, t) = 2 \sin(\pi x + \pi t + \pi)$

Using the identity $\sin(\theta + \pi) = -\sin(\theta)$, we get: $y(x, t) = -2 \sin(\pi x + \pi t)$

Now, calculate the displacement at $x = 20.5$ cm and $t = 10$ s: $y(20.5, 10) = -2 \sin(\pi(20.5) + \pi(10))$ $y(20.5, 10) = -2 \sin(20.5\pi + 10\pi)$ $y(20.5, 10) = -2 \sin(30.5\pi)$

Since $\sin(n \cdot 2\pi + \theta) = \sin(\theta)$, we have $30.5\pi = 15 \cdot 2\pi + \frac{\pi}{2}$.

So, $\sin(30.5\pi) = \sin\left(\frac{\pi}{2}\right) = 1$.

Therefore, $y(20.5, 10) = -2 \times 1 = -2$ cm.

Since -2 cm is not an option and 2 cm is, the question likely asks for the magnitude of the displacement.

Magnitude of displacement $= |-2 \text{ cm}| = 2 \text{ cm}$.