MedVision ad

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

Sugar

Member
Joined
Dec 18, 2003
Messages
163
Location
Sydney
Gender
Female
HSC
2005
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

Member
Joined
Nov 1, 2003
Messages
299
Gender
Male
HSC
2003
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

Member
Joined
Dec 18, 2003
Messages
163
Location
Sydney
Gender
Female
HSC
2005
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

Member
Joined
Jan 30, 2005
Messages
276
Location
Sydney
Gender
Male
HSC
2005
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
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top