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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

Two monoatomic ideal gases 1 and 2 has molecular weights \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\). Both are kept in two different containers at the same temperature. The ratio of velocity of sound wave in gas 1 and 2 is $\ldots \ldots \ldots$ (A) \(\sqrt{\left(m_{2} / m_{1}\right)}\) (B) \(\sqrt{\left(m_{1} / m_{2}\right)}\) (C) \(\left(\mathrm{m}_{1} / \mathrm{m}_{2}\right)\) (D) \(\left(\mathrm{m}_{2} / \mathrm{m}_{1}\right)\)

If the equation of a wave in a string having linear mass density $0.04 \mathrm{~kg} \mathrm{~m}^{-1}\( is given by \)\mathrm{y}=0.02\( \)\sin [2 \pi\\{1 /(0.04)\\}-\\{\mathrm{x} /(0.50)\\}]$, then the tension in the string is \(\ldots \ldots \ldots \ldots\) N. (All values are in \(\mathrm{mks}\) ) (A) \(6.25\) (B) \(4.0\) (C) \(12.5\) (D) \(0.5\)

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) \(1050 \mathrm{~Hz}\)

As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and $\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}$ and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).When the block is moving towards \(k_{1}\), what will be the time taken for it to get maximum compressed from point \(\mathrm{C}\) ? (A) \(\pi \mathrm{s}\) (B) \((2 / 3) \mathrm{s}\) (C) \((\pi / 3) \mathrm{s}\) (D) \((\pi / 4) \mathrm{s}\)

The tension in a wire is decreased by \(19 \%\), then the percentage decrease in frequency will be....... (A) \(19 \%\) (B) \(10 \%\) (C) \(0.19 \%\) (D) None of these
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