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Chapter 10: Problem 1439

If the resultant of two waves having amplitude \(\mathrm{b}\) is \(\mathrm{b}\), then the phase difference between the two waves is (A) \(120^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)

Short Answer

Expert verified
The phase difference between the two waves is \(90^{\circ}\).

Step by step solution

01

Write down the superposition equation for two waves with amplitude b and phase difference φ

The superposition equation for the resultant of two waves with amplitude b and a phase difference of φ is: \[R = \sqrt{b^2 + b^2 + 2bb \cos(\phi)}\]
02

Solve the equation for R=b

Since we know that R = b, we can substitute it in the equation and solve for φ to find the phase difference. \[b = \sqrt{b^2 + b^2 + 2bb \cos(\phi)}\]
03

Solve for the phase difference φ

We will now simplify the equation and solve for φ: \[b^2 = b^2 + b^2 + 2b^2 \cos(\phi)\] \[0 = b^2 + 2b^2 \cos(\phi)\] Using the property for cos(θ) : \[\cos(90^{\circ}- \phi)=-\cos(\phi)\] Divide by b^2 on each side of the equation: \[0 = 1 + 2 \cos(\phi)\] Now, substitute the new property of cos(θ) in the equation, \[0 = 1 - 2\cos(90^{\circ} - \phi)\]
04

Discard the possible multiple solutions

The above equation can have multiple solutions for the value of φ, as the equation depends on cos(θ). However, it is important to consider that the phase difference must be within the range from 0° to 180°.
05

Find the correct option

Comparing our equation with the given options, we find a match with the option (C): (C) \(90^{\circ}\) Therefore, the phase difference between the two waves is \(90^{\circ}\).

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Most popular questions from this chapter

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of amplitudes $\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots$ (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

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) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

A string of length \(70 \mathrm{~cm}\) is stretched between two rigid supports. The resonant frequency for this string is found to be \(420 \mathrm{~Hz}\) and \(315 \mathrm{~Hz}\). If there are no resonant frequencies between these two values, then what would be the minimum resonant frequency of this string ? (A) \(10.5 \mathrm{~Hz}\) (B) \(1.05 \mathrm{~Hz}\) (C) \(105 \mathrm{~Hz}\) (D) \(1050 \mathrm{~Hz}\)

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