Quick SHM Question

  • #1
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Hello,

So I've been doing old practice questions on SHM to revise, and just been frustrating myself on this one for a bit. The question is:

"An object vibrating with simple harmonic motion has a maximum speed of 1.6 m s−1 and maximum magnitude of the acceleration of 8π m s−2. Calculate the period of the motion. "

I can't remember why, but I've divided the speed by 2*pi and then divided that by the acceleration [(1.6*2*pi) / 8*pi], and it's given me the correct answer of 0.40s but I can't quite understand why that's correct! Lol

Cheers,
Niall
 
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Answers and Replies

  • #2
gneill
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What relevant equations do you know that pertain to SHM? In particular, what equation is often used to write the position as a function of time?
 
  • #3
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What relevant equations do you know that pertain to SHM? In particular, what equation is often used to write the position as a function of time?
Do you mean: x(t) = Acos(ωt + φ)?
 
  • #4
gneill
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Do you mean: x(t) = Acos(ωt + φ)?
Sure. That'll work. Now, given that function for displacement, how might you find velocity?
 
  • #5
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Sure. That'll work. Now, given that function for displacement, how might you find velocity?
By differentiating, giving: v(t) = Aωsin(ωt + φ), and acceleration would therefore be: a(t) = -Aω2cos(ωt + φ)
 
  • #6
gneill
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Good, good. So you know how to move from displacement to velocity to acceleration through differentiating the function.

Since you're given a value for maximum speed, say V = 1.6 m/s, then your velocity function might be v(t) = V sin(ωt + φ), right? Can you work from there?
 
  • #7
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Good, good. So you know how to move from displacement to velocity to acceleration through differentiating the function.

Since you're given a value for maximum speed, say V = 1.6 m/s, then your velocity function might be v(t) = V sin(ωt + φ), right? Can you work from there?
I got to that point previously, but just wasn't sure what to do next. I feel like it should be obvious, but it just hasn't clicked
 
  • #8
gneill
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I got to that point previously, but just wasn't sure what to do next. I feel like it should be obvious, but it just hasn't clicked
Differentiate the velocity expression...
 
  • #9
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Differentiate the velocity expression...
Ok, dv(t) / dt = -Vωcos(ωt + φ)
 
  • #10
gneill
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What's the maximum value of the result?
 
  • #11
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What's the maximum value of the result?
I would think that is when the phase is equal to zero, giving cos(ωt + φ) = 1
 
  • #12
gneill
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I would think that is when the phase is equal to zero, giving cos(ωt + φ) = 1
Sure. And what is the maximum value? What multiplies the cosine function?
 
  • #13
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Sure. And what is the maximum value? What multiplies the cosine function?
dv(t)/dt = -Vω
 
  • #14
gneill
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Ok, dv(t) / dt = -Vωcos(ωt + φ)
I should have caught this earlier: the derivative of sin is cos, not -cos. Not that it'll affect the eventual result, but I thought I should point it out.
dv(t)/dt = -Vω
The magnitude of the value that multiplies the trig function is the important thing: Since sin and cos both range between +1 and -1, its the magnitude of the value multiplying the sin or cos that sets the maximum value of the function. In this case it's the maximum value of the acceleration, right? So don't write dv(t)/dt = -Vω since you've dropped the trig function and it's no longer a function of time. It's the maximum value. Instead, write A = Vω, where A is the maximum acceleration...
 
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  • #15
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I should have caught this earlier: the derivative of sin is cos, not -cos. Not that it'll affect the eventual result, but I thought I should point it out.

The magnitude of the value that multiplies the trig function is the important thing: Since sin and cos both range between +1 and -1, its the magnitude of the value multiplying the sin or cos that sets the maximum value of the function. In this case it's the maximum value of the acceleration, right? So don't write dv(t)/dt = -Vω since you've dropped the trig function and it's no longer a function of time. It's the maximum value. Instead, write A = Vω, where A is the maximum acceleration...
Oh of course! Bloody hell. Yeah, thank you for going through that and having patience. I should have got this a hell of a lot easier, but it feels like one of those nights lol, thanks very much! Appreciate it.
 
  • #16
gneill
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No worries. Glad to have helped.
 
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