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A disc of radius \(20 \mathrm{~m}\) is rotating uniformly with angular frequency \(\omega=10 \mathrm{rad} / \mathrm{s}\), A source is fixod to rim of dise. The ratio of maximum and minimum froquency heard by observer far away from disc in plane of disc is (take speed of sound \(330 \mathrm{~m} / \mathrm{s}\) ) (a) \(33 / 13\) (b) \(33 / 53\) (c) \(13 / 53\) (d) \(53 / 13\)

Short Answer

Expert verified
The answer is (d) \( \frac{53}{13} \).

Step by step solution

01

Identify Parameters

First, note the parameters given in the problem: the radius of the disc is \( R = 20 \, \mathrm{m} \), the angular frequency is \( \omega = 10 \, \mathrm{rad/s} \), and the speed of sound \( c = 330 \, \mathrm{m/s} \). The frequency of the source is not specified, but we are interested in the ratio of the maximum to the minimum frequency heard by an observer.
02

Understand Doppler Effect in Rotating System

In a rotating system, the Doppler effect will depend on the velocity of the source with respect to the observer. The maximum and minimum frequencies are reached when the source is moving directly towards or directly away from the observer.
03

Determine Linear Velocity of Source

The linear velocity \( v_s \) of the source can be determined using \( v_s = \omega R \). Substituting the given values: \[ v_s = 10 \, \mathrm{rad/s} \times 20 \, \mathrm{m} = 200 \, \mathrm{m/s} \].
04

Apply Doppler Effect Equations

Calculate frequencies heard using the Doppler effect equations. For maximum frequency \( f_+ = \frac{f_0 (c + v_s)}{c} \) and for minimum frequency \( f_- = \frac{f_0 (c - v_s)}{c} \), where \( f_0 \) is the original frequency.
05

Compute Ratio of Maximum and Minimum Frequencies

Since we are interested in the ratio:\[ \text{Ratio} = \frac{f_+}{f_-} = \frac{c + v_s}{c - v_s} = \frac{330 + 200}{330 - 200} = \frac{530}{130} = \frac{53}{13} \].
06

Select Correct Answer

The calculated ratio of maximum to minimum frequency \( \frac{53}{13} \) matches option (d) in the given choices.

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Key Concepts

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

Rotational Motion Physics
Rotational motion physics deals with the motion of objects that rotate around a fixed point or axis. This kind of motion is everywhere, from clock hands to the wheels of a car. In this context, a disc is an excellent example,
as it spins around its center. This spinning is not just a linear motion; it involves a combination of several factors that make it unique.
  • Angular Position: It describes the orientation of a line with another line or plane, simply a measure of the angle.
  • Angular Displacement: This is the change in the angular position when the object rotates.
  • Angular Velocity: Denoted by \(\omega\), it's the rate of change of angular displacement.
  • Angular Acceleration: This measures how quickly the angular velocity changes.
The unique aspect of rotational motion is how it relates to linear motion through the radius of the disc. For instance, if a point is moving around the disc, its velocity depends on both the angular velocity and its distance from the center, which is the radius. This way, many of the formulas for linear motion have rotational counterparts, such as \( v = \omega R \), where \( v \) is the linear velocity.
Angular Frequency
Angular frequency, represented by \(\omega\), quantifies how fast something rotates or oscillates. It's a measure often used in systems that follow periodic motion, such as pendulums or rotating wheels.
To understand angular frequency, we can look at its relation to the concept of frequency that measures cycles per second, denoted by \( f \). The difference is that while regular frequency is in cycles per second (Hertz), angular frequency is in radians per second.
  • Key Formula: \( \omega = 2\pi f \)
  • In our example, \(\omega\) is provided as \(10\, \mathrm{rad/s}\), meaning the disc completes \(10/2\pi\) full rotations each second.
  • This rate of rotation helps determine the linear velocity of points on the rotating object.
Angular frequency is essential for analyzing motions like the spreading ripples in water or the pulsating rotations of galaxies. It's the bridge that helps convert linear properties to circular ones, like understanding how fast car tires spin based on engine RPMs.
Speed of Sound
The speed of sound is a crucial aspect when studying wave phenomena such as the Doppler effect. In any medium, it's affected by several factors, including temperature and the medium itself—such as air, water, or metal.
In this exercise, we're considering sound in air, which typically moves at about \(330 \, \mathrm{m/s}\). This speed is close to its value at room temperature (20°C). Sound travels in waves, and understanding this helps us predict how quickly it will reach an observer.
  • In a rotating system, like our disc, the speed of sound determines how the frequency will change due to motion.
  • The observer perceives sound differently based on whether the sound source is moving toward them or away from them.
  • The Doppler effect precisely uses this change in frequency to understand how fast the source or observer is moving relative to each other.
The significance of the speed of sound becomes apparent as it influences everything from daily weather reports to cutting-edge scientific experiments or technologies like radar and sonar.

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Most popular questions from this chapter

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(a) Let \(L \rightarrow\) stationary, \(\nu=580 \mathrm{II} \mathrm{z}\) \(s \rightarrow\) moving with \(40 \mathrm{~km} / \mathrm{hr}\) and aproaching \(w \rightarrow 40 \mathrm{~km} / \mathrm{hr}\) and supporting \(v=1200 \mathrm{~km} / \mathrm{hr}\) So \(v^{\prime}=\left[\frac{v+\omega}{(v+\omega)-v_{s}}\right] v\) \(\Rightarrow \quad v^{\prime}=\frac{1240}{1200} \times 580=\frac{1798}{3}=599.33 \mathrm{IIz}\) (b) At a distance \(1 \mathrm{~km}\) before the train whistles and let the driver heard the echo at time \(l=t^{\prime}\). then at this time, total distance travelled by wave and train \(=2 \mathrm{~km}\) Now time after which first wave reach to the hill in time interval \(t_{1}=\frac{1}{1200} \mathrm{hr}\) after reflecting at \(t=t^{\prime}\), echo will heard, then distance travellcd by it is \(x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200\) and distance travelled by the train is \(x_{2}=\left(t^{\prime}-\frac{1}{1200}\right) 40\) As \(x_{1}+x_{2}=1 \mathrm{~km}\) \(\therefore\left(t^{\prime}-\frac{1}{1200}\right) \times 12040+\left(t^{\prime}-\frac{1}{1200}\right) 40\) \(1240 t^{\prime}=\frac{1}{30} \quad \therefore \quad t^{\prime}=\frac{61}{30 \times 1240}\) and required distance, $$ x_{1}=\left(t^{\prime}-\frac{1}{1200}\right) \times 1200 $$$=\left(\frac{61}{30 \times 1240}-\frac{1}{1200}\right) \times 1200=\frac{61 \times 1200}{30 \times 1240}-1\( \)=\frac{30}{31} \times 1000 \mathrm{~m}=967.74 \mathrm{~m}\( Frequency of sound heard by the driver, \)v^{\prime \prime}=\left(\frac{(v-\omega)+v_{2}}{v-\omega}\right) v\( \)=\left(\frac{1200-40+40}{1200-40}\right) \times 599.33\( \)\therefore \quad v^{\prime \prime}=\frac{1200}{1160} \times 599.33=620 \mathrm{~Hz}$

Statement-1: Oscillations in spring mass system are due to continuous exchange of energy between the block and the spring. Statcment-2 : The total energy of the spring mass system remains constant.

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