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Chapter 9: Problem 3

Two SHMs are represented by the equations \(Y_{1}=10\) \(\sin \left(3 \pi t+\frac{\pi}{4}\right)\) and \(Y_{2}=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\) Their amplitudes are in the ratio of (A) \(2: 1\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 4\)

Short Answer

Expert verified
The amplitudes of the given SHMs are 10 for \(Y_1 = 10 \sin(3\pi t + \frac{\pi}{4})\) and 5 for \(Y_2 = 5\sin(3\pi t + \tan^{-1}(\sqrt{3}))\). The ratio of their amplitudes is \(\frac{10}{5} = 2\), which gives a ratio of 2:1. The correct answer is (A).

Step by step solution

01

Identify the equations of the SHMs

We are given the equations of two SHMs, which are: \(Y_1 = 10 \sin(3\pi t + \frac{\pi}{4})\) and \(Y_2 = 5(\sin(3\pi t) + \sqrt{3} \cos(3\pi t))\)
02

Find the amplitude of each SHM

The amplitude of a SHM is the maximum value that can be attained by the function representing the motion. For \(Y_1\), the amplitude is given by the coefficient of the sine function, which is 10. For \(Y_2\), let's first rewrite the equation using the trigonometric identity: \(\sin\alpha \cos\beta + \cos\alpha \sin\beta = \sin(\alpha + \beta)\) We can let \(\alpha = 3\pi t\) and \(\beta = \tan^{-1}(\sqrt{3})\). Then, we have: \(Y_2 = 5(\sin(3\pi t)\cos(\tan^{-1}(\sqrt{3})) + \sqrt{3} \cos(3\pi t)\sin(\tan^{-1}(\sqrt{3})))\) Using the trigonometric identity, we get: \(Y_2 = 5\sin(3\pi t + \tan^{-1}(\sqrt{3}))\) Now, the amplitude of \(Y_2\) is the coefficient of the sine function, which is 5.
03

Determine the ratio of the amplitudes

We have found the amplitudes of \(Y_1\) and \(Y_2\), which are 10 and 5, respectively. The ratio of their amplitudes is: \(\frac{10}{5} = 2\) Therefore, the ratio of the amplitudes of SHMs \(Y_1\) and \(Y_2\) is 2:1. The correct answer is (A).

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Most popular questions from this chapter

A wave travelling along the \(x\)-axis is described by the equation \(y(x, t)=0.005 \cos (\alpha x-\beta t)\). If the wavelength and the time period of the wave are \(0.08 \mathrm{~m}\) and \(2.0 \mathrm{~s}\), respectively, then \(\alpha\) and \(\beta\) in appropriate units are (A) \(\alpha=25.00 \pi, \beta=\pi\) (B) \(\alpha=\frac{0.08}{\pi}, \beta=\frac{2.0}{\pi}\) (C) \(\alpha=\frac{0.04}{\pi}, \beta=\frac{1.0}{\pi}\) (D) \(\alpha=12.50 \pi, \beta=\frac{\pi}{2.0}\)

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A particle of mass \(m\) moves in a straight line. If \(v\) is the velocity at a distance \(x\) from a fixed point on the line and \(v^{2}=a-b x^{2}\), where \(a\) and \(b\) are constant, then (A) The motion continues along the positive \(x\)-direction only. (B) The motion is simple harmonic. (C) The particle oscillates with a frequency equal to \(\frac{\sqrt{b}}{2 \pi}\) (D) The total energy of the particle is \(m a\).

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