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Chapter 11: Problem 33

Using graph paper, sketch two identical sine waves of amplitude $4.0 \mathrm{cm}\( that differ in phase by (a) \)\pi / 3$ rad \(\left(60^{\circ}\right)\) and (b) \(\pi / 2\) rad \(\left(90^{\circ}\right) .\) Find the amplitude of the superposition of the two waves in each case.

Short Answer

Expert verified
Answer: The amplitudes of the superpositions for the two cases are approximately 6.93 cm and 5.66 cm, respectively.

Step by step solution

01

Sketch the first sine wave

Begin by drawing a sinusoidal curve that represents the first sine wave. This sine wave should have an amplitude of 4 cm.
02

Sketch the second sine wave (60掳 phase difference)

Now, draw another sine wave that has an amplitude of 4 cm and a phase difference of \(\pi / 3\) rad (60掳) following the first sine wave.
03

Calculate the amplitude of the superposition

To find the amplitude of the superposition of the two sine waves, we can use the formula: \(A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta\phi)}\), where \(A\) is the amplitude of the superposition, \(A_1\) and \(A_2\) are the amplitudes of the initial waves, and \(\Delta\phi\) is the phase difference between the waves. In our case, \(A_1 = A_2 = 4 \thinspace cm\), and \(\Delta\phi = \pi / 3\). So, \(A = \sqrt{4^2 + 4^2 + 2(4)(4) \cos(\pi / 3)} = \sqrt{48} \thinspace cm \approx 6.93 \thinspace cm\). #b) pi/2 (90 degrees) phase difference#
04

Sketch the sine wave (90掳 phase difference)

With the same amplitude of 4 cm, draw another sine wave that has a phase difference of \(\pi / 2\) rad (90掳) from the first sine wave.
05

Calculate the amplitude of the superposition

Now we will use the same formula to find the amplitude of the superposition but with \(\Delta\phi = \pi / 2\): \(A = \sqrt{4^2 + 4^2 + 2(4)(4) \cos(\pi / 2)} = \sqrt{32} \thinspace cm \approx 5.66 \thinspace cm\).

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