How do you find maximum acceleration of a particle? (1 Viewer)

Sugar

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Q:
A particleis moving in SHM. Its displacement x at time t is give by
x = 3sin(2t + 5)

Find maximum acceleration of the particle.

Thanks in advance.
 

toknblackguy

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ooh ooh i know this one ;)

differentiate twice to get to acc
..
x = -12 sin(2t + 5)

and now looking only at the sin(2t+5) part, this can only have a max value of 1 (it range of values of from -1 to 1)

so max acc is 12
 

Sugar

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Oh, that seems simple! It was in Q2 of 1997 HSC Extension 1 paper, and I was kicking myself because I couldn't do it.

So, the same would apply to cos, yeah?

Thankyou! :D
 

Ghost1788

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Another way of doing this question..

Because x=3sin(2t+5) is in the form x=asin(nt+k)
Where a=3 and n=2
And Since the motion is in Simple Harmonic Motion
..
x = -(n^2)*x

..
x = -4x

therefore max acceleration is at x = -3

Therefore max acceleration = -4 * -3 = 12
 

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