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Chapter 16: Problem 6

In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is \(0.170 \mathrm{~s}\). The frequency of the wave is (a) \(1.47 \mathrm{~Hz}\) (b) \(0.36 \mathrm{~Hz}\) (c) \(0.73 \mathrm{~Hz}\) (d) \(2.94 \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of the wave is (a) 1.47 Hz.

Step by step solution

01

Understanding Sinusoidal Wave Movement

A sinusoidal wave oscillates between a maximum displacement (peak) and zero displacement (equilibrium) in a quarter of its full period because it takes one full period to go through the process of maximum displacement, zero, minimum, and back to maximum. Therefore, the time from maximum to zero is one-fourth of its period, denoted as \( T/4 \).
02

Calculating the Period from Given Time

We are given that moving from maximum displacement to zero displacement takes \( 0.170 \text{ s} \). Therefore, the full period \( T \) of the wave is \( 4 \times 0.170 \text{ s} \). Calculating this gives us \( T = 0.680 \text{ s} \).
03

Determine Frequency from Period

Frequency \( f \) is the reciprocal of the period \( T \). Therefore, we calculate \( f \) using the formula: \( f = \frac{1}{T} = \frac{1}{0.680} \text{ Hz} \). Calculating this gives: \( f \approx 1.47 \text{ Hz} \).
04

Identifying the Correct Answer from Options

From the previous calculation, we found that the frequency \( f \approx 1.47 \text{ Hz} \). Looking at the choices given, the frequency that matches our calculation is option (a) \( 1.47 \text{ Hz} \).

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Key Concepts

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

Sinusoidal Wave
A sinusoidal wave is a smooth, repetitive oscillation that typically describes the motion of waves in a variety of contexts, such as sound waves, light waves, and water waves. The term "sinusoidal" is derived from the mathematical sine function, which graphically represents the wave's motion:
  • It has a regular pattern of crests and troughs.
  • The wave completes one cycle when it returns to a starting point, going through maximum, zero, and minimum displacements.
  • It's characterized by its amplitude (maximum height), wavelength (distance between successive crests), and frequency (number of cycles per second).
In analyzing sinusoidal waves, understanding the movement from maximum displacement to zero helps in determining various properties, like the wave's period and frequency. This motion is crucial as it represents only a quarter of the full oscillation cycle.
Oscillation Period
The oscillation period of a wave is defined as the time it takes to complete one full cycle. For sinusoidal waves, this cycle includes the motion from maximum displacement through zero to minimum displacement and back to maximum.
  • It is denoted by the letter "T" and measured in seconds.
  • The given problem specifies that the wave takes 0.170 seconds to go from maximum to zero displacement.
  • Since this represents just a quarter of the wave's full period, the total period is calculated as four times that time: \( T = 4 \times 0.170 \text{ s} = 0.680 \text{ s} \).
The period is inversely related to the wave鈥檚 frequency: if the period increases, the frequency decreases, and vice versa.
Maximum Displacement
Maximum displacement in a sinusoidal wave refers to the highest point, or apex, the wave reaches during its oscillation.
  • This point is often referred to as the "peak" or "crest" of the wave.
  • It represents the amplitude of the wave, which is a measure of how much energy the wave is carrying.
  • A wave moves from its maximum displacement to zero, indicating energy is moving through the medium.
Understanding this transition helps in solving for key wave properties like period and frequency, as knowing the time to move from maximum displacement to equilibrium can aid in calculating the full period and, subsequently, the wave's frequency.

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Most popular questions from this chapter

Two forks \(A\) and \(B\) when sounded together produce four beats \(\mathrm{s}^{-1}\). The fork \(A\) is in unison with \(30 \mathrm{~cm}\) length of a sonometer wire and \(B\) is in unison with \(25 \mathrm{~cm}\) length of the same wire at the same tension. The frequencies of the forks are (a) \(24 \mathrm{~Hz}, 28 \mathrm{~Hz}\) (b) \(20 \mathrm{~Hz}, 24 \mathrm{~Hz}\) (c) \(16 \mathrm{~Hz}, 20 \mathrm{~Hz}\) (d) \(26 \mathrm{~Hz}, 30 \mathrm{~Hz}\)

\(v_{1}\) and \(v_{2}\) are the velocities of sound at the same temperature in two monoatomic gases of densities \(\rho_{1}\) and \(\rho_{2}\) respectively. If \(\frac{\rho_{1}}{\rho_{2}}=\frac{1}{4}\), then the ratio of velocities \(v_{1}\) and \(v_{2}\) will be (a) \(1: 2\) (b) \(4: 1\) (c) \(2: 1\) (d) \(1: 4\)

Oxygen is 16 times heavier than hydrogen. Equal volumes of hydrogen and oxygen are mixed. The ratio of speed of sound in the mixture to that in hydrogen is (a) \(\sqrt{\frac{1}{8}}\) (b) \(\sqrt{\frac{32}{17}}\) (c) \(\sqrt{8}\) (d) \(\sqrt{\frac{2}{17}}\)

Four simple harmonic vibrations, $$ \begin{aligned} &y_{1}=8 \cos \omega t \\ &y_{2}=4 \cos \left(\omega t+\frac{\pi}{2}\right) \\ &y_{3}=2 \cos (\omega t+\pi) \\ &y_{4}=\cos \left(\omega t+\frac{3 \pi}{2}\right) \end{aligned} $$ are superimposed on one another. The resulting amplitude and phase are respectively (a) \(\sqrt{45}\) and \(\tan ^{-1}\left(\frac{1}{2}\right)\) (b) \(\sqrt{45}\) and \(\tan ^{-1}\left(\frac{1}{3}\right)\) (c) \(\sqrt{75}\) and \(\tan ^{-1}(2)\) (d) \(\sqrt{75}\) and \(\tan ^{-1}\left(\frac{1}{3}\right)\)

A cylindrical tube open at both ends, has a fundamental frequency \(f_{0}\) in air. The tube is dipped vertically into water such that half of its length inside water. The fundamental frequency of the air column now is (a) \(3 f_{0} / 4\) (b) \(f_{0}\) (c) \(f_{0} / 2\) (d) \(2 f_{0}\)
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