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Chapter 15: Problem 10

Two points are located at a distance of \(10 \mathrm{~m}\) and \(15 \mathrm{~m}\) from the source of oscillation. The period of oscillation is \(0.05 \mathrm{~s}\) and the velocity of the wave is \(300 \mathrm{~m} / \mathrm{s}\). What is the phase difference between the oscillations of two points? (a) \(\pi\) (b) \(\frac{\pi}{6}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{2 \pi}{3}\)

Short Answer

Expert verified
The phase difference between the oscillations is \( \frac{2\pi}{3} \).

Step by step solution

01

Calculate the Wavelength

The wavelength \( \lambda \) of a wave can be calculated using the formula: \( \lambda = \frac{v}{f} \), where \( v \) is the velocity of the wave and \( f \) is the frequency. First, we calculate the frequency using the period \( T \): \( f = \frac{1}{T} = \frac{1}{0.05} = 20 \ \text{Hz} \). Now we can find the wavelength: \( \lambda = \frac{300}{20} = 15 \ \text{m} \).
02

Calculate the Path Difference

The path difference \( \Delta x \) between the two points is the difference in their distances from the source, which is \( 15 \ \text{m} - 10 \ \text{m} = 5 \ \text{m} \).
03

Calculate the Phase Difference

The phase difference \( \Delta \phi \) is calculated using the formula: \( \Delta \phi = \frac{2\pi \cdot \Delta x}{\lambda} \). Substituting the known values: \( \Delta \phi = \frac{2\pi \cdot 5}{15} = \frac{2\pi}{3} \).

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Key Concepts

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

Phase Difference
When waves travel through different distances from their source, they might not arrive in phase, meaning they don't match up in their wave cycles. The phase difference is the amount by which one wave leads or lags behind another. This is measured in radians or degrees. Soon, you will grasp why phase differences are important in understanding how waves interact and interfere with each other.

Knowing the
  • path difference: the actual difference in distance traveled by the waves
  • and wavelength helps us calculate the phase difference.
The phase difference formula is: \[\Delta \phi = \frac{2\pi \cdot \Delta x}{\lambda} \]This connects the path difference \( \Delta x \) with the phase difference \( \Delta \phi \). It shows how many wave cycles fit into the path difference.

For example, if the calculated phase difference is
  • \( \frac{2\pi}{3} \), it indicates that one wave is 2/3 of a cycle ahead of the other.
  • This suggests constructive or destructive interference with other waves nearby depending on their phase relationship.
Wavelength Calculation
The wavelength, denoted \( \lambda \), is a crucial aspect of wave mechanics. It represents the length of one complete wave cycle, including one crest and one trough. Understanding how to calculate wavelength is essential to resolving many problems related to waves.

To calculate the wavelength, we use the formula:\[\lambda = \frac{v}{f} \]where:
  • \( v \) is the velocity of the wave.
  • \( f \) is the frequency of oscillation.
By knowing the speed at which the wave travels and how many waves pass a certain point per second, we can determine the distance one wave covers. This relationship is key to understanding how waves move and interact in real-world scenarios.
Frequency of Oscillation
The frequency of oscillation refers to how frequently the wave peaks pass a particular point in a given time. It's measured in Herz (Hz), which is equivalent to cycles per second. The frequency is inversely proportional to the period of oscillation, which is the time it takes for one complete wave cycle.

The formula used is:\[f = \frac{1}{T} \]where \( T \) is the period in seconds.

This implies:
  • If \( T \) is small, meaning each cycle takes less time, the frequency is high.
  • If \( T \) is large, the frequency is low, meaning fewer cycles occur each second.
Understanding frequency is crucial for gaining insights into how fast different waves oscillate and interact with their environment, thus influencing wave interference and other phenomena it can cause. This understanding is not only theoretical but has applications ranging from sound waves to light, and even radio signals.

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Most popular questions from this chapter

A pendulum is made to hang from a ceilling of an elevator.It has period of \(T_{\mathrm{sec}}\). (for small angles). The elevator is made to accelerate upwards with \(10 \mathrm{~m} / \mathrm{s}^{2}\).The period of the pendulum now will be (assume \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(T \sqrt{2}\) (b) infinite (c) \(T / \sqrt{2}\) (d) zero

Motion of an oscillating liquid column in a \(U\)-tube is [NCERT Exemplar] (a) periodic but not simple harmonic (b) non-periodic (c) simple harmonic and time period is independent of the density of the liquid (d) simple harmonic and time-period is directly proportional to the density of the liquid

A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is \(T\). With what acceleration should the lift be accelerated upwards in order to reduce its period to T/2? ( \(g\) is the acceleration due to gravity) [Karnataka CET 2008] (a) \(4 \mathrm{~g}\) (b) \(g\) (c) \(2 \mathrm{~g}\) (d) \(3 \bar{g}\)

A mass \(1 \mathrm{~kg}\) suspended from a spring whose force constant is \(400 \mathrm{Nm}^{-1}\), executes simple harmonic oscillation. When the total energy of the oscillator is \(2 \mathrm{~J}\), the maximum acceleration experienced by the mass will be (a) \(2 \mathrm{~ms}^{-2}\) (b) \(4 \mathrm{~ms}^{-2}\) (c) \(40 \mathrm{~ms}^{-2}\) (d) \(400 \mathrm{~ms}^{-2}\)

The particle execute simple harmonic motion with a time period of \(16 \mathrm{~s}\). At time \(t=2 \mathrm{~s}\), the particle crosses the mean position while at \(t=4 \mathrm{~s}\), its velocity is \(4 \mathrm{~ms}^{-1} .\) The amplitude of motion in metre is (a) \(\sqrt{2} \pi\) (b) \(16 \sqrt{2} \pi\) (c) \(32 \sqrt{2 /} \pi\) (d) \(4 / \pi\) (e) \(24 \sqrt{2} \pi\)
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