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. 2 cm
• B. -2 cm
• C. 0 cm
• D. 4 cm

Answer 1

Answer: (B)

-2 cm

Answer 2

To determine the displacement of the particle, we first need to establish the equation of the sinusoidal wave.

1. General Equation of the Wave:

A sinusoidal wave traveling in the negative x-direction can be represented by the equation:

$y(x, t) = A \sin(kx + \omega t + \phi)$

where:

• $A$ is the amplitude
• $k$ is the wave number
• $\omega$ is the angular frequency
• $\phi$ is the phase constant

2. Given Parameters:

• Amplitude, $A = 2$ cm
• Distance between two consecutive crests (wavelength), $\lambda = 2$ cm
• Angular frequency, $\omega = \pi$ s$^{-1}$

3. Calculate Wave Number (k):

The wave number $k$ is related to the wavelength $\lambda$ by the formula:

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

Substituting the given wavelength:

$k = \frac{2\pi}{2 \text{ cm}} = \pi \text{ cm}^{-1}$

4. Substitute Known Values into the Wave Equation:

Now, the wave equation becomes:

$y(x, t) = 2 \sin(\pi x + \pi t + \phi)$

5. Determine the Phase Constant ($\phi$):

We are given the initial condition: at $t = 0$ and $x = 0$, the particle was at its mean position ($y = 0$) moving downward.

• Condition 1: $y(0, 0) = 0$

  Substitute $x = 0$ and $t = 0$ into the wave equation:

  $0 = 2 \sin(\pi(0) + \pi(0) + \phi)$

  $0 = 2 \sin(\phi)$

  $\sin(\phi) = 0$

  This implies $\phi = n\pi$, where $n$ is an integer (e.g., $0, \pi, 2\pi, ...$).

• Condition 2: Moving downward at $x = 0, t = 0$

  This means the velocity of the particle, $\frac{\partial y}{\partial t}$, is negative at $x = 0, t = 0$.

  First, find the expression for the velocity by differentiating $y(x, t)$ with respect to $t$:

  $\frac{\partial y}{\partial t} = \frac{\partial}{\partial t} [2 \sin(\pi x + \pi t + \phi)]$

  $\frac{\partial y}{\partial t} = 2 \cos(\pi x + \pi t + \phi) \cdot (\pi)$

  $\frac{\partial y}{\partial t} = 2\pi \cos(\pi x + \pi t + \phi)$

  Now, apply the condition at $x = 0, t = 0$:

  $(\frac{\partial y}{\partial t})_{x=0, t=0} = 2\pi \cos(\pi(0) + \pi(0) + \phi) < 0$

  $2\pi \cos(\phi) < 0$

  $\cos(\phi) < 0$

  Combining $\sin(\phi) = 0$ and $\cos(\phi) < 0$:

  • If $\phi = 0$, $\sin(0) = 0$ but $\cos(0) = 1$ (not less than 0).
  • If $\phi = \pi$, $\sin(\pi) = 0$ and $\cos(\pi) = -1$ (which is less than 0).

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

6. Complete Wave Equation:

Substitute $\phi = \pi$ back into the wave equation:

$y(x, t) = 2 \sin(\pi x + \pi t + \pi)$

Using the trigonometric identity $\sin(\theta + \pi) = -\sin(\theta)$:

$y(x, t) = -2 \sin(\pi x + \pi t)$

7. Calculate Displacement at $x = 20.5$ cm and $t = 10$ s:

Substitute these values into the derived wave equation:

$y(20.5, 10) = -2 \sin(\pi(20.5) + \pi(10))$

$y(20.5, 10) = -2 \sin(20.5\pi + 10\pi)$

$y(20.5, 10) = -2 \sin(30.5\pi)$

We can write $30.5\pi$ as $30\pi + 0.5\pi = 30\pi + \frac{\pi}{2}$.

Since $\sin(2n\pi + \theta) = \sin(\theta)$ for any integer $n$:

$\sin(30\pi + \frac{\pi}{2}) = \sin(\frac{\pi}{2})$

We know $\sin(\frac{\pi}{2}) = 1$.

So, $y(20.5, 10) = -2 \times 1$

$y(20.5, 10) = -2$ cm

The displacement of the particle at $x = 20.5$ cm and $t = 10$ s is -2 cm.