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A small source of sound oscillates in simple harmonic motion with an amplitude of \(17 \mathrm{~cm}\). A detector is placed along the line of motion of the source. The source emits a sound of frequency \(800 \mathrm{~Hz}\) which travels at a speed of \(340 \mathrm{~m} \mathrm{~s}^{-1}\). If the width of the frequency band detected by the detector is \(8 \mathrm{~Hz}\), find the time period of the source.

Short Answer

Expert verified
The time period of the source is approximately 0.63 seconds.

Step by step solution

01

Understand the Problem

We have a sound source oscillating in simple harmonic motion (SHM) with known amplitude, frequency, and speed of sound. The width of the frequency band detected is the difference between the maximum and minimum frequencies observed by the detector due to the Doppler effect. We need to find the time period of this source, which is oscillating in SHM.
02

Establish Equations for Doppler Effect

The observed frequency by the detector can vary between the maximum and minimum values due to the Doppler effect. Using the formula for frequency observed due to the Doppler effect in the case of a moving source: \( f' = \frac{f \cdot (v)}{v \pm v_s}, \)where \(f'\) is the observed frequency, \(f\) the emitted frequency, \(v\) the speed of sound, and \(v_s\) the source velocity. The frequency band is \(f_{max} - f_{min} = 8\, \text{Hz}.\)
03

Calculate Source Velocity at Maximum Displacement

In SHM, the maximum speed of the source occurs at the equilibrium position and is given by:\[v_s = \omega A,\]where \(\omega\) is the angular frequency and \(A\) is the amplitude (17 cm = 0.17 m). Convert amplitude to meters and use it to find \(\omega\).
04

Relate Frequency Band to Maximum Source Velocity

The width of the frequency band detected,\(8\, \text{Hz}\), corresponds to the difference when the source moves towards and away from the detector. At maximum speed:\[f_{max} - f_{min} = \frac{2f \cdot v_s}{v}. \]Substitute given values and solve for \(v_s\).
05

Compute Angular Frequency and Period

From the relation found between the source velocity and displacement, solve for angular frequency:\[v_s = \omega A.\]Since \(\omega = \frac{2\pi}{T}\), solve for \(T\):\[T = \frac{2\pi A}{v_s}.\]
06

Solve for Time Period T

Using the expressions derived, substitute \(A = 0.17\, \text{m}\) and calculated \(v_s\) from Step 4 into:\[T = \frac{2\pi \cdot 0.17}{v_s}.\] Find \(v_s\) from previous steps, and calculate \(T\). Assume previously calculated \(v_s = \frac{8 \times 340}{2 \times 800} = 1.7\, \text{m/s},\) so:\[T = \frac{2\pi \times 0.17}{1.7} \approx 0.63\, \text{s}. \]

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Key Concepts

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

Doppler Effect
The Doppler Effect is a fundamental concept in physics, describing the change in frequency or wavelength of a wave in relation to an observer moving relative to the wave source. This effect is commonly experienced in everyday life, such as when a passing vehicle sounds differently as it approaches and then moves away.

In the context of sound waves, the Doppler Effect explains how the detected frequency at a stationary point can change when the sound source is in motion. The formula \( f' = \frac{f \cdot v}{v \pm v_s} \) helps to calculate the observed frequency \( f' \), where \( f \) is the original frequency of the source, \( v \) is the speed of sound, and \( v_s \) is the velocity of the source.

The effect causes the frequency band observed by a detector, which is the range between the maximum and minimum frequencies, when a source moves in simple harmonic motion. This change in frequency can be used to calculate various properties of the source's motion, such as its time period.
Frequency Band
A frequency band refers to the range of frequencies over which an oscillating system emits or responds. In simple harmonic motion, when a sound source moves, the frequency detected by a stationary observer changes due to the Doppler Effect. The difference between the maximum and minimum frequencies observed is known as the width of the frequency band.

In our example, the emitted frequency is \( 800 \mathrm{~Hz} \), and the width of the frequency band detected is \( 8 \mathrm{~Hz} \). This width represents how much the frequency shifts as the source moves toward and away from the detector. Understanding the frequency band is crucial in analyzing how quickly or slowly the source is moving and helps provide insight into the dynamics of the system in question.
Angular Frequency
Angular Frequency, noted as \( \omega \), is an important parameter in oscillatory motion, especially in simple harmonic motion (SHM). It represents how fast an object oscillates in a circular path, measured in radians per second. It is closely related to the concept of frequency but specifically deals with the angular displacement over time.

The relationship between angular frequency and frequency \( f \) is given by the equation \( \omega = 2\pi f \). In SHM, angular frequency is key to determining other aspects like the source's velocity; it can be calculated using the formula \( \omega = \frac{v_s}{A} \), where \( v_s \) is the source velocity and \( A \) is the amplitude.

Angular frequency is integral for solving problems involving oscillations, such as calculating the maximum speed of the source or its time period, as it provides a connection between linear motion parameters and circular motion metrics.
Time Period Calculation
The time period \( T \) is the duration it takes for one complete cycle of oscillation, a significant parameter in analyzing oscillatory systems. It is inversely related to the frequency of the source. The relationship between the angular frequency and the time period is defined by \( T = \frac{2\pi}{\omega} \).

In solving time period problems, one must often determine the source velocity at maximum displacement within its simple harmonic motion. Given this velocity and amplitude, the angular frequency can be calculated, which then leads to the determination of the time period. In the provided solution, the time period is calculated using \( T = \frac{2\pi A}{v_s} \), where \( A \) is the amplitude and \( v_s \) is the known source velocity.

Understanding how to calculate the time period from parameters like amplitude and velocity equips learners with tools to solve a variety of oscillation-related problems efficiently.

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Most popular questions from this chapter

Two coherent narrow slits emitting sound of wavelength \(\lambda\) in the same phase are placed parallel to each other at a small separation of \(2 \lambda\). The sound is detected by moving a detector on the screen \(\Sigma\) at a distance \(D(\gg>)\) from the slit \(S_{1}\) as shown in figure (16-E6). Find the distance \(x\) such that the intensity at \(P\) is equal to the intensity at \(O\).

A person standing on a road sends a sound signal to the driver of a car going away from him at a speed of \(72 \mathrm{~km} \mathrm{~h}^{-1}\). The signal travelling at \(330 \mathrm{~m} \mathrm{~s}^{-1}\) in air and having a frequency of \(1600 \mathrm{~Hz}\) gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person.

Ultrasonic waves of frequency \(4 \cdot 5 \mathrm{MHz}\) are used to detect tumour in soft tissues. The speed of sound in tissue is \(15 \mathrm{~km} \mathrm{~s}^{-1}\) and that in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\). Find the wavelength of this ultrasonic wave in air and in tissue.

An operator sitting in his base camp sends a sound signal of frequency \(400 \mathrm{~Hz}\). The signal is reflected back from a car moving towards him. The frequency of the reflected sound is found to be \(410 \mathrm{~Hz}\). Find the speed of the car. Speed of sound in air \(=324 \mathrm{~m} \mathrm{~s}^{-1}\).

A tuning fork of frequency \(256 \mathrm{~Hz}\) produces 4 beats per second with a wire of length \(25 \mathrm{~cm}\) vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?

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