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Chapter 10: Problem 1351

The displacement of a S.H.O. is given by the equation \(\mathrm{x}=\mathrm{A}\) \(\cos \\{\omega t+(\pi / 8)\\}\). At what time will it attain maximum velocity? (A) \((3 \pi / 8 \omega)\) (B) \((8 \pi / 3 \omega)\) (C) \((3 \pi / 16 \omega)\) (D) \((\pi / 16 \pi)\).

Short Answer

Expert verified
The time at which the S.H.O. will attain its maximum velocity is (C) \(\frac{3 \pi}{16 \omega}\).

Step by step solution

01

Identify the given equation

We have the displacement equation for a S.H.O.: x = A cos(ωt + π/8) where A is the amplitude, ω is the angular frequency, and t is the time.
02

Differentiate the displacement equation

In order to find the velocity equation, we need to differentiate the displacement equation with respect to time: \(v = \frac{dx}{dt} = \frac{d(A\cos(\omega t + \frac{\pi}{8}))}{dt}\) Now, apply the chain rule for the differentiation: \(v = -A\omega \sin(\omega t + \frac{\pi}{8})\)
03

Determine the maximum value of the velocity equation

The maximum value of the velocity equation is when \(\sin{(\omega t + \frac{\pi}{8})}\) is equal to 1, its maximum value: \(\sin{(\omega t + \frac{\pi}{8})} = 1\) Now we need to discover at what time the sine part becomes equal to 1.
04

Find the time at which the maximum velocity is attained

To find the time at which \(\sin{(\omega t + \frac{\pi}{8})} = 1\), we consider the inverse sine function with 1: \(\omega t + \frac{\pi}{8} = \sin^{-1}(1)\) Since the inverse sine of 1 is π/2 (or every odd multiple of π/2), we can write: \(\omega t + \frac{\pi}{8} = \frac{\pi (2n+1)}{4}\) , where n is an integer. Now, solve for t: \(t = \frac{1}{\omega}\left(\frac{\pi (2n+1)}{4} - \frac{\pi}{8}\right)\) \(t = \frac{3 \pi(2n+1)}{16\omega}\) By checking the options, we notice that option (C) corresponds to the case when n=0:
05

Choose the correct answer

The time at which the S.H.O. will attain its maximum velocity is: (C) \(\frac{3 \pi}{16 \omega}\)

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