A2 SHM Physics question Watch

CurrentDude
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#1
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This is a question on SHM.
I've difficulty in solving it. Here is the question:

The displacement of an object oscillating in SHM varies with time in accordance with the equation x/mm = 12 Cos10t, where t is the time in seconds after the object's displacement was at its maximum positive value.

a. Determine
(i) amplitude (ii) the time period

b. Calculate the displacement of the object at i = 0.1 s

Can anyone solve it? Please also show all workings

Cheers
CD
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TSR Jessica
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Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses.

I'm going to quote in Tank Girl now so she can move your thread to the right place if it's needed. :yy:

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(Original post by Tank Girl)
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Joinedup
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well

the cos of any real number can only ever take values between +1 and -1 inclusive... so the value of 12 cos (anything) must be between +12 and -12 inclusive.

if you drew a graph with cos(x) on the y axis it'd be a 'sine wave' shape that crossed the x=0 axis at (0,+1) and repeated every 2π... it'd have a positive peak every time x/2π was a whole number

the reason π appears is because for SHM type questions, radians are the conventional (and most natural) units of measurement.
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CurrentDude
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Now I've difficulty in following your answer..... . both parts (a) & (b) require numbers !!! ???
and you've not stated that.
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Joinedup
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(Original post by CurrentDude)
Now I've difficulty in following your answer..... . both parts (a) & (b) require numbers !!! ???
and you've not stated that.
You can type 12 cos 10t into google and the first thing on the page is the graph you should be sketching if you got this question in an exam. https://www.google.co.uk/search?q=12+cos+10t

displacement is on the y axis and time is on the x axis.

can you see what the amplitude is?

the period is the amount of time between one peak and the next, we're told that the argument to cos is 10t and we know cos (x) has peaks when x is 0,2π,4π,6π etc...

so the period is just 2π/10
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Jpw1097
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(Original post by CurrentDude)
This is a question on SHM.
I've difficulty in solving it. Here is the question:

The displacement of an object oscillating in SHM varies with time in accordance with the equation x/mm = 12 Cos10t, where t is the time in seconds after the object's displacement was at its maximum positive value.

a. Determine
(i) amplitude (ii) the time period

b. Calculate the displacement of the object at i = 0.1 s

Can anyone solve it? Please also show all workings

Cheers
CD
The general equation for SHM is x = Acos(wt), where A = amplitude, and w=2pi/T.

So for part (i) the amplitude should be pretty straight forward to work out.

As for part (ii), since w=10 ==> 2pi/T=10. Hopefully you can work out the time period from there.

For part b, you simply need to plug the time into the function to find displacement.

Hope that helps!
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CurrentDude
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(Original post by Jpw1097)
The general equation for SHM is x = Acos(wt), where A = amplitude, and w=2pi/T.

So for part (i) the amplitude should be pretty straight forward to work out.

As for part (ii), since w=10 ==> 2pi/T=10. Hopefully you can work out the time period from there.

For part b, you simply need to plug the time into the function to find displacement.

Hope that helps!
Thanx for the solution. I got the amplitude. And I also came to the same answer for the rest. Solved it this afternoon. I'll post my answer soon so that others might find it useful.

Cheers
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