Question

Consider two points lying at a distance of \(10 \mathrm{~m}\) and \(15 \mathrm{~m}\) from an oscillating source. If the periodic time of oscillation is \(0.05 \mathrm{~s}\) and the velocity of wave produced is \(300 \mathrm{~m} / \mathrm{s}\), then what will be the phase difference the two points? (A) \(\pi\) (B) \(\pi / 6\) (C) \(\pi / 3\) (D) \(2 \pi / 3\)

Short answer

The phase difference between the two points is \(\frac{2\pi}{3}\), so the correct answer is (D) \(\frac{2\pi}{3}\).

Step-by-step solution

Determine the frequency of the wave

To determine the phase difference, we first need to find the frequency of the wave produced by the oscillating source. Given the periodic time of oscillation \(T\), the frequency \(f\) can be calculated using the formula: \[ f = \frac{1}{T} \]

Calculate the wavelength of the wave

Now we need to find the wavelength of the wave. We can use the wave speed formula, which relates frequency, wavelength, and wave velocity: \[ v = f\lambda \] Where \(v\) is the wave velocity, \(f\) is the frequency, and \(\lambda\) is the wavelength. We can solve this equation for the wavelength: \[ \lambda = \frac{v}{f} \]

Determine the path difference between the two points

The path difference between two points is simply the difference in distances from the oscillating source. In this case: \[ \Delta x = 15 \, m - 10 \, m = 5 \, m \]

Calculate the phase difference

Now we can determine the phase difference between the two points using the path difference and wavelength. The phase difference \(\Delta \phi\) can be calculated using the formula: \[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x \] Substitute the values obtained in Steps 1, 2, and 3 into the formula: 1. Calculate the frequency \[ f = \frac{1}{0.05 \, s} = 20 \, Hz \] 2. Calculate the wavelength \[ \lambda = \frac{300 \, m / s}{20 \, Hz} = 15 \, m \] 3. Calculate the phase difference \[ \Delta \phi = \frac{2\pi}{15 \, m} \cdot 5 \, m = \frac{2\pi}{3} \] The phase difference between the two points is \(\frac{2\pi}{3}\), so the correct answer is (D) \(\frac{2\pi}{3}\).

Key concepts

Frequency Calculation in Waves

Understanding the frequency of a wave is essential because it helps us analyze how often certain oscillatory events occur over time. The concept is quite simple: frequency refers to the number of complete wave cycles passing a point per second. A higher frequency means more wave cycles per second. For instance, if we have a frequency of 20 Hz, this means 20 wave cycles pass a point in one second.

We calculate wave frequency (\(f\)) using the formula \( f = \frac{1}{T} \), where \( T \) is the periodic time of oscillation. The periodic time is the duration of one complete wave cycle. In our problem, the periodic time given is 0.05 seconds. The calculation would be:

• \( f = \frac{1}{0.05 \, \text{s}} = 20 \, \text{Hz} \)

By understanding frequency, we can better comprehend various types of wave behaviors in different mediums.

Wavelength Determination Formula

The wavelength is a crucial property of any wave, as it represents the physical length of one complete cycle of the wave. In essence, it measures the distance between two consecutive corresponding points in the wave, such as from crest to crest. In our scenario, we used the wave speed to find this value.

Waves travel with a velocity, which can be represented by the equation \( v = f\lambda \), linking wave speed \( v \), frequency \( f \), and wavelength \( \lambda \). By rearranging this equation to solve for \( \lambda \), we have:

• \( \lambda = \frac{v}{f} \)

Given the wave velocity (\(v = 300 \, \text{m/s}\)) and frequency (\(f = 20 \, \text{Hz}\)), you can calculate the wavelength:

• \( \lambda = \frac{300 \, \text{m/s}}{20 \, \text{Hz}} = 15 \, \text{m} \)

Recognizing and calculating wavelength helps us analyze patterns in wave propagation.

Oscillation Mechanics

Oscillation mechanics explain the periodic movement seen in waves, where particles of the medium vibrate back and forth around an equilibrium point. This regular movement is crucial for understanding processes such as sound or light propagation, where energy transfers without the overall shift of medium particles.

Waves produced by oscillating sources exhibit properties such as amplitude, frequency, and phase. The frequency, as already discussed, is the rate of oscillation, while the amplitude indicates the maximum extent of movement. The concept of phase is also fundamental, as it shows how aligned an oscillating particle is with a standard point in a wave cycle. Understanding how these parameters interact helps us comprehend complex physical phenomena.

This knowledge becomes vital when analyzing two points' behavior at differing distances from an oscillating source and calculating the impact on wave phase.

Wave Velocity Concepts

Wave velocity is the speed at which a wave travels through a medium. It is an essential attribute, impacting the design of technology such as radio, television, and internet communications. The velocity describes how quickly oscillatory energy passes through space.

In our problem scenario, we use the formula for wave velocity \( v = f\lambda \), which helps connect the frequency and wavelength of a wave with its speed. Knowing the speed and how to calculate it: comes in handy for solving practical problems related to waves.

For instance, understanding that the wave produced by the oscillating source travels at 300 m/s, allows us to determine other aspects, such as the alignment or phase between points separated by a certain distance. This concept helps us in mapping out how wave behaviors change based on various speeds.