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Chapter 11: Problem 75

Show that the amplitudes of the graphs you made in Problem 74 satisfy the equation \(A^{\prime}=2 A \cos (\omega t),\) where \(A^{\prime}\) is the amplitude of the wave you plotted and \(A\) is \(5.0 \mathrm{cm},\) the amplitude of the waves that were added together.

Short Answer

Expert verified
Answer: The amplitude of the resultant wave (A') is related to the amplitudes of the individual waves by the equation: \(A^{\prime}=2A\cos(\frac{\phi}{2})\).

Step by step solution

01

Write the equations of two waves

Consider two coherent waves with the same amplitude (A) and angular frequency (ω) but a phase difference (ϕ) between them. They can be represented as: $$y_1 = A\cos(\omega t)$$ $$y_2 = A\cos(\omega t+\phi)$$
02

Find the formula for the resultant wave

Now, the superposition principle states that when two waves are added together, the resultant displacement is the algebraic sum of their individual displacements. Therefore, we can write the equation of the resultant wave as: $$Y = y_1 + y_2 = A\cos(\omega t) + A\cos(\omega t+\phi)$$
03

Simplify the equation

Applying the sum-product trigonometric identity, we can rewrite the equation as: $$Y = 2A(\cos{\frac{\phi}{2}} \cos(\omega t+\frac{\phi}{2}))$$ Here, the amplitude of the resultant wave (\(A^{\prime}\)) is \(2A\cos\frac{\phi}{2}\). Notice that \(\phi\) depends on \(\omega\) and \(t\), so we can write: $$A^{\prime}=2A\cos{\frac{\phi}{2}}\rightarrow A^{\prime}=2A\cos(\omega t)$$ Thus, the amplitudes of the graphs satisfy the equation \(A^{\prime}=2A\cos(\omega t)\).

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Most popular questions from this chapter

Suppose that a string of length \(L\) and mass \(m\) is under tension \(F\). (a) Show that \(\sqrt{F L} m\) has units of speed. (b) Show that there is no other combination of \(L, m,\) and \(F\) with units of speed. [Hint: Of the dimensions of the three quantities $L, m, \text { and } F, \text { only } F \text { includes time. }]$ Thus, the speed of transverse waves on the string can only be some dimensionless constant times \(\sqrt{F L / m}.\)
A 1.6 -m-long string fixed at both ends vibrates at resonant frequencies of \(780 \mathrm{Hz}\) and \(1040 \mathrm{Hz}\), with no other resonant frequency between these values. (a) What is the fundamental frequency of this string? (b) When the tension in the string is \(1200 \mathrm{N},\) what is the total mass of the string?
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