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Chapter 14: Problem 115

A deepwater wave of wavelength \(\lambda\) has a speed given approximately by \(v=\sqrt{g \lambda / 2 \pi}\). Find an expression for the period of a deepwater wave in terms of its wavelength. (Note the similarity of your result to the period of a pendulum.)

Short Answer

Expert verified
The period \(T\) is \(\frac{2\pi \sqrt{\lambda}}{\sqrt{g}}\).

Step by step solution

01

Identify Given Formula

The speed of a deepwater wave is given by the formula: \(v = \sqrt{\frac{g \lambda}{2 \pi}}\), where \(v\) is the speed, \(g\) is the acceleration due to gravity, and \(\lambda\) is the wavelength.
02

Recall the Relationship Between Speed, Wavelength, and Period

For a wave, the speed \(v\) is related to its wavelength \(\lambda\) and period \(T\) by the equation: \(v = \frac{\lambda}{T}\). This formula states that speed is equal to wavelength divided by the period.
03

Substitute the Expression for Speed

Substitute the expression for speed from Step 1 into the wave speed formula: \[\frac{\lambda}{T} = \sqrt{\frac{g \lambda}{2 \pi}}\].
04

Manipulate the Equation to Solve for Period \(T\)

Start by squaring both sides to eliminate the square root: \[\left(\frac{\lambda}{T}\right)^2 = \frac{g \lambda}{2 \pi}\]. Next, rewrite the left side: \[\frac{\lambda^2}{T^2} = \frac{g \lambda}{2 \pi}\].
05

Solve for \(T^2\)

Multiply both sides by \(T^2\) and divide by \(g \lambda\) to isolate \(T^2\): \[T^2 = \frac{4 \pi^2 \lambda}{g}\].
06

Solve for \(T\)

Take the square root of both sides to solve for the period \(T\): \[T = \frac{2\pi \sqrt{\lambda}}{\sqrt{g}}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed Formula
The wave speed formula is crucial in understanding how waves propagate through different mediums. For deepwater waves, the formula used is:\[ v = \sqrt{\frac{g \lambda}{2 \pi}} \]This formula shows that the speed of a wave \(v\) depends on the gravitational acceleration \(g\) and the wavelength \(\lambda\). In this context:
  • \(v\) (speed): This is the rate at which the wave travels through water.
  • \(g\) (gravity): Acceleration due to Earth's gravity, approximately \(9.81 \text{ m/s}^2\).
  • \(\lambda\) (wavelength): The distance between two consecutive crests or troughs of the wave.
By using this equation, we can determine how variations in gravity and wavelength affect the wave speed. It highlights a direct proportional relationship between the wave's speed and its wavelength, meaning that longer waves travel faster when gravity is constant.
Wavelength
Wavelength is a key component when studying waves. It is the length from one point on a wave to the identical point on the next wave, typically measured from crest to crest or trough to trough. This can be visualized as the horizontal distance covered by one complete cycle of the wave.Understanding wavelength is essential as it directly influences wave speed and period:
  • Longer Wavelengths: Higher wave speeds and longer periods.
  • Shorter Wavelengths: Lower wave speeds and shorter periods.
In the case of deepwater waves, the speed and period can be derived from the wavelength using the wave speed formula and the relationship between speed, wavelength, and period:\[ v = \frac{\lambda}{T} \]This equation reveals that as wavelength increases, wave speed increases, affecting the wave's period as well.
Gravity
Gravity plays a fundamental role in wave dynamics, especially for deepwater waves. It is the force that pulls objects towards the Earth, influencing how waves behave on the surface of the water.For deepwater waves, gravity impacts:
  • Wave Speed: Waves travel faster in deeper water scenarios where gravity is a significant factor.
  • Wave Period: The period of the wave, or the time it takes for a complete cycle of the wave to pass a point, is affected by gravitational acceleration.
In the wave speed formula, gravity is denoted by \(g\), approximately \(9.81 \text{ m/s}^2\). It shows how gravitational forces help determine the propagation speed of the wave along with its wavelength:\[ v = \sqrt{\frac{g \lambda}{2 \pi}} \]This relationship emphasizes the essential nature of gravity in calculating and predicting the behavior of waves in oceans and large bodies of water.

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Most popular questions from this chapter

Hearing a Pin Drop The ability to hear a "pin drop" is the sign of sensitive hearing. Suppose a \(0.55-g\) pin is dropped from a height of \(28 \mathrm{cm},\) and that the pin emits sound for \(1.5 \mathrm{s}\) when it lands. Assuming all of the mechanical energy of the pin is converted to sound energy, and that the sound radiates uniformly in all directions, find the maximum distance from which a person can hear the pin drop. (This is the ideal maximum distance, but atmospheric absorption and other factors will make the actual maximum distance considerably smaller.)

A steel guitar string has a tension \(T\), length \(L\), and diameter D. Give the multiplicative factor by which the fundamental frequency of the string changes under the following conditions: (a) The tension in the string is increased by a factor of 4 The diameter is \(D\) and the length is \(L\). (b) The diameter of the string is increased by a factor of \(3 .\) The tension is \(T\) and the length is \(L\). (c) The length of the string is halved. The tension is \(T\) and the diameter is \(D\).

When you blow across the top of a soda pop bottle you hear a fundamental frequency of \(206 \mathrm{Hz}\). Suppose the bottle is now filled with helium. (a) Does the fundamental frequency increase, decrease, or stay the same? Explain. (b) Find the new fundamental frequency. (Assume that the speed of sound in helium is three times that in air.)

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