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simple harmonic motion (1 Viewer)

Smilebuffalo

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The period of a particle moving in SHM is 6 s and its amplitude is 8cm. Calculate its velocity and acceleration (correct to 1 decimal place) when the displacement is 5cm from the centre of motion.

:)

answer: v = -6.5cm/s a = -5.5cm/s/s
 

scardizzle

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Period = 6

therefore 2pi/n = 6

therefore n = pi/3

a = 8

using the general form v^2 = n^2(a^2 - x^2)

v^2 = (pi/3)^2(8^2 - 5^2)

= 42.768..

therefore v = +/- 6.5397... cm/s not sure why answer is negative though...

a = -n^2x

= - (pi/3)^2 x 5

= - 5.4831 cm/s
 

annabackwards

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Period = 6

therefore 2pi/n = 6

therefore n = pi/3

a = 8

using the general form v^2 = n^2(a^2 - x^2)

v^2 = (pi/3)^2(8^2 - 5^2)

= 42.768..

therefore v = +/- 6.5397... cm/s not sure why answer is negative though...

a = -n^2x

= - (pi/3)^2 x 5

= - 5.4831 cm/s
Naww, you beat me to it ^^
 

Timothy.Siu

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well u know period=2pi/n

and a=-n^2 . x

v^2=n^2(a^2-x^2)

so first, lets find n,

6=2pi/n
n=2pi/6

when x=5, a=-(2pi/6)^2 . 5 =5.48311....=5.5 (1d.p)

v^2=n^2(64-25)
v^2=39n^2
v^2 = 42.76828....
v=+-6.5 depending which way it is going at this time.(i'm not really sure about this, coz i got this wrong in my exam lol)
 

Smilebuffalo

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coool thanks guys..

I was just a bit confused because in this question i didn't know what the centre of motion was so i didn't know if i should use either:

i) v^2 = n^2(a^2 - x^2)

or

ii) v^2 = n^2(a^2 - (x-b)^2)


And also i'm confused too about when i'm meant to choose between a positive and negative velocity. :S
 
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scardizzle

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coool thanks guys..

I was just a bit confused because in this question i didn't know what the centre of motion was so i didn't know if i should use either:

i) v^2 = n^2(a^2 - x^2)

or

ii) v^2 = n^2(a^2 - (x-b)^2)


And also i'm confused too about when i'm meant to choose between a positive and negative velocity. :S
1.since the questions specifies its 5m away from its centre of motion you can ignore b

2.usually you work out whether to take the positive/negative root when the questions gives you a result e.g. the question states when x = 3 v = -5

you would put this into your equation and see whether the negative or positive root is required

I dont think there is enough information in the question for us to tell
 

Drongoski

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Period = 6



therefore v = +/- 6.5397... cm/s not sure why answer is negative though...


When particle heads to the right (+ve direction), velocity is +ve

and when it heads in the opposite direction (to the left). v is -ve.


PS: 5 cm from centre can be 5cm to its right or 5cm to its left.

In each cycle(1 period):


Particle passes thru 5cm to the right twice, once on its way to the right extremity (v +ve) and once on its way back towards the left extremity (v -ve). Similarly, for 5cm to the left, particle passes thru this point twice (v -ve and then +ve)

So if I've not misread question, answer for v should be both + and - 6.5 . . .
 
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