Question

A sinusoidal wave has an amplitude of 3 cm, a time period of 1 s, and a wavelength of 1 cm. At x = 0 and t = 0, the particle is at its mean position and moving upward. What is the equation of the wave if it is propagating in the positive x-direction?

• A. y = 3cos($\pi$ + 2$\pi$t - 2$\pi$x)
• B. y = 3sin(2$\pi$t - 2$\pi$x)
• C. y = 3sin(2$\pi$x - 2$\pi$t)
• D. x = 3sin(2$\pi$t - 2$\pi$y)

Answer 1

Answer: (B)

y = 3sin(2$\pi$t - 2$\pi$x)

Answer 2

The general equation of a sinusoidal wave propagating in the positive x-direction is given by:

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

where:

$A$ is the amplitude

$k$ is the wave number ($k = \frac{2\pi}{\lambda}$)

$\omega$ is the angular frequency ($\omega = \frac{2\pi}{T}$)

$\phi$ is the phase constant

Given values:

Amplitude, $A = 3$ cm

Time period, $T = 1$ s

Wavelength, $\lambda = 1$ cm

1. Calculate the wave number ($k$):

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

2. Calculate the angular frequency ($\omega$):

$\omega = \frac{2\pi}{T} = \frac{2\pi}{1 \text{ s}} = 2\pi \text{ rad/s}$

3. Substitute A, k, and $\omega$ into the general wave equation:

$y(x, t) = 3 \sin(2\pi x - 2\pi t + \phi)$

4. Determine the phase constant ($\phi$) using the initial conditions:

At $x = 0$ and $t = 0$, the particle is at its mean position ($y = 0$).

$0 = 3 \sin(2\pi(0) - 2\pi(0) + \phi)$

$0 = 3 \sin(\phi)$

$\sin(\phi) = 0$

This implies $\phi = n\pi$, where $n$ is an integer. Possible values are $0, \pi, 2\pi, ...$.

At $x = 0$ and $t = 0$, the particle is moving upward. This means its velocity, $\frac{\partial y}{\partial t}$, is positive.

First, find the expression for the velocity:

$\frac{\partial y}{\partial t} = \frac{\partial}{\partial t} [3 \sin(2\pi x - 2\pi t + \phi)]$

$\frac{\partial y}{\partial t} = 3 \cos(2\pi x - 2\pi t + \phi) \cdot (-2\pi)$

$\frac{\partial y}{\partial t} = -6\pi \cos(2\pi x - 2\pi t + \phi)$

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

$(\frac{\partial y}{\partial t})_{x=0, t=0} = -6\pi \cos(2\pi(0) - 2\pi(0) + \phi) > 0$

$-6\pi \cos(\phi) > 0$

$\cos(\phi) < 0$

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

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

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

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

5. Write the complete wave equation:

$y(x, t) = 3 \sin(2\pi x - 2\pi t + \pi)$

6. Simplify the equation using trigonometric identities:

We know that $\sin(\theta + \pi) = -\sin(\theta)$.

So, $y(x, t) = -3 \sin(2\pi x - 2\pi t)$

We also know that $-\sin(\theta) = \sin(-\theta)$.

So, $y(x, t) = 3 \sin(-(2\pi x - 2\pi t))$

$y(x, t) = 3 \sin(2\pi t - 2\pi x)$

Therefore, the correct answer is y = 3sin(2$\pi$t - 2$\pi$x).