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Chapter 10: Problem 1433

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

Expert verified
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

01

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: \[ f = \frac{1}{T} \]
02

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: \[ \lambda = \frac{v}{f} \]
03

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: \[ \Delta x = 15 \, m - 10 \, m = 5 \, m \]
04

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) \(\frac{2\pi}{3}\).

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Most popular questions from this chapter

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

If \((1 / 4)\) of a spring having length \(\ell\) is cutoff, then what will be the spring constant of remaining part? (A) \(\mathrm{k}\) (B) \(4 \mathrm{k}\) (C) \((4 \mathrm{k} / 3)\) (D) \((3 \mathrm{k} / 4)\)

A body of mass \(1 \mathrm{~kg}\) suspended from the free end of a spring having force constant \(400 \mathrm{Nm}^{-1}\) is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is $\ldots \ldots . \mathrm{ms}^{-2}$. (A) \(8 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-2}\) (C) \(40 \mathrm{~ms}^{-2}\) (D) \(40 \mathrm{cms}^{-2}\)
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