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Chapter 10: Problem 1346

The amplitude for a S.H.M. given by the equation $\mathrm{x}=3 \sin 3 \mathrm{pt}+4 \cos 3 \mathrm{pt}\( is \)\ldots \ldots \ldots \ldots \mathrm{m}$ (A) 5 (B) 7 (C) 4 (D) \(3 .\)

Short Answer

Expert verified
The amplitude for the given SHM equation is \(A = 5\,m\).

Step by step solution

01

Write down the given equation in general SHM form

The given equation for the SHM is \(x = 3\sin(3pt) + 4\cos(3pt)\). We aim to rewrite this equation in the general SHM form: \(x(t) = A\sin(\omega t + \phi)\).
02

Apply the sum-to-product trigonometric identity

We use the sum-to-product trigonometric identity to convert the sum of sine and cosine functions into a product form: \[\sin(A) \cos(B) + \cos(A) \sin(B) = \sin(A + B)\] Using this identity, we rewrite our given equation as follows: \[x = R\sin(3pt + \phi)\] where \(R = \sqrt{3^2 + 4^2}\) and \(\tan{\phi} = \frac{3}{4}\).
03

Calculate R

Calculate the value of R using the Pythagorean theorem: \[R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
04

Identify the amplitude

The amplitude, A, is equal to the value of R in our transformed equation: \[A = R = 5\]
05

Find the correct option

The correct option for the amplitude is: (A) 5

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Most popular questions from this chapter

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

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 tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before? (A) \(480 \mathrm{~Hz}\) (B) \(490 \mathrm{~Hz}\) (C) \(460 \mathrm{~Hz}\) (D) \(470 \mathrm{~Hz}\)

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 frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)
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