Question

The velocity of sound in sea water is about 1530 m/sec. Write an equation for a sinusoidal sound wave in the ocean, of amplitude 1 and frequency 1000 hertz.

Short answer

\( y(t) = \sin(2\pi \times 1000 \, t - 4.11x) \)

Step-by-step solution

Understand the Given Information

The problem states that the velocity of sound in sea water is 1530 m/sec. The amplitude of the sound wave is 1, and the frequency is 1000 hertz. We need to write an equation for a sinusoidal sound wave.

Write the General Form of a Sinusoidal Wave

The general form of a sinusoidal wave can be written as \( y(t) = A \, \sin(2\pi f t - kx) \), where \( A \) is the amplitude, \( f \) is the frequency, \( t \) is time, \( k \) is the wave number, and \( x \) is the position.

Substitute the Given Amplitude

We know the amplitude \( A \) is 1. Therefore, \( y(t) = \sin(2\pi f t - kx) \).

Substitute the Given Frequency

The frequency \( f \) is 1000 hertz. So the equation becomes: \( y(t) = \sin(2\pi \times 1000 \, t - kx) \).

Calculate the Wave Number \( k \)

The wave number \( k \) is given by \( k = \frac{2\pi}{\lambda} \), where \( \lambda \) is the wavelength. From the wave speed equation \( v = f \lambda \), we have \( \lambda = \frac{v}{f} \). Here, \( v = 1530 \) m/sec and \( f = 1000 \) hertz, so \( \lambda = \frac{1530}{1000} = 1.53 \) m. Thus, \( k = \frac{2\pi}{1.53} \).

Substitute the Wave Number \( k \)

Now, substitute \( k \) into the equation: \( k = \frac{2\pi}{1.53} \approx 4.11 \) So the final equation is: \( y(t) = \sin(2\pi \times 1000 \, t - 4.11x) \).

Key concepts

Sinusoidal Wave

A sinusoidal wave is a mathematical curve that describes a smooth, periodic oscillation. It's similar to the shape of a sine or cosine function and is commonly seen in various physical phenomena. In our context, a sinusoidal sound wave in the ocean can be visualized as a wave that oscillates up and down smoothly over time. These waves can describe things like sound, light, and even the oscillation of a swinging pendulum.
The general form of a sinusoidal wave can be written as
\[ y(t) = A \, \sin(2\pi f t - kx) \]
where:

• \(A\) is the amplitude, indicating the maximum displacement from the mean position.
• \(f\) is the frequency, showing how many cycles the wave completes in one second.
• \(t\) is the time variable.
• \(k\) is the wave number.
• \(x\) is the spatial variable along the direction of wave propagation.

Sound Velocity

The velocity of sound (or sound speed) in a medium like sea water is the speed at which sound waves propagate through that medium. In our problem, the velocity of sound in sea water is given as 1530 meters per second (m/s). Sound travels faster in water than in air due to water's higher density. This value is essential as it affects the wavelength and consequently the wave number. For a given frequency, the speed of sound determines how 'stretched' or 'compressed' each wave cycle appears.
We use the relationship
\[ v = f \lambda \]
where:

• \(v\) is the sound velocity.
• \(f\) is the frequency of the sound wave.
• \(\lambda\) is the wavelength.

This equation helps us calculate the wavelength from the known frequency and sound velocity.

Wave Number

The wave number is a measure of the number of wavelengths per unit distance. It's represented by the symbol \(k\) and relates to the wavelength through the equation:
\[ k = \frac{2\pi}{\lambda} \]
The wave number gives us an idea of how many wave cycles fit into a unit distance along the propagation direction. In our problem, to find \(k\), we first determined the wavelength \(\lambda\) from the velocity and frequency:

• \( \lambda = \frac{v}{f} \)
  where \(v = 1530\) m/s and \(f = 1000\) Hz.
  This gives \( \lambda = 1.53 \) meters.
• Next, we calculate \( k \) using \( \lambda \):
  \( k = \frac{2\pi}{1.53} \approx 4.11 \)

Thus, our final wave equation included this wave number value.

Frequency

Frequency in the context of waves defines how many complete cycles or oscillations occur in one second. It's measured in hertz (Hz). For our sinusoidal sound wave in the ocean, the problem states a frequency of 1000 Hz. High frequency means more cycles per second, leading to a higher pitch if you think about sound.
The frequency is directly used in our wave equation as part of the sine function argument:
\[ y(t) = \sin(2\pi \times 1000 \, t - kx) \]
Here, \(2\pi f t\) indicates how the wave oscillates over time.

Amplitude

Amplitude refers to the maximum extent of a wave's displacement from its mean (rest) position. It indicates how 'strong' or 'intense' the wave is. For the sinusoidal sound wave in the ocean given in this problem, the amplitude is 1. This means the peak (or trough) of the wave reaches a value of 1.
The amplitude appears directly in the wave equation:
\[ y(t) = A \, \sin(2\pi f t - kx) \]
Here, \(A = 1\), forming:
\[ y(t) = \sin(2\pi \times 1000 \, t - 4.11x) \]
The amplitude determines the height of the sine wave but does not affect its frequency or wavelength.