An oscillating object very easy question (I think)

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  • #1
s3a
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"An oscillating object" very easy question (I think)

Sorry if this is a really stupid question but if it is then it'll be easy to answer . For the attached of example 15.1, I don't get how to do letter d. For v_max for instance, I see that sin(whateverIsInside) should equal -1 because -4pi*(-1) = 4pi. My work is attached.

Any help would be greatly appreciated!
Thanks in advance!
 

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Answers and Replies

  • #2
diazona
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Think about this first: when you have a function
[tex]f(t) = C\sin(\text{something})[/tex]
where C is some constant, what is the maximum possible value that function can have?

By the way, for future reference, it would be nice to at least summarize the question, and include your work, in the post itself, in addition to attaching PDF files.
 
  • #3
s3a
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Sorry, you are right, I will summarize it next time if I have another question. As for your question the answer is +/- C. I think I am confused with the inside of the sine function.
 
  • #4
diazona
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Actually +C would be the maximum. -C would be the minimum. (Unless C is a negative constant, then they'd be switched) But you basically have the right idea. The point is that the argument of the sine function (the "something" in my last post) doesn't even matter. You were able to identify the maximum (and minimum) without knowing anything about what was inside the sin(). Try applying that reasoning to your problem.
 
  • #5
s3a
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Yes, but that only explains it to me mathematically. I want to understand the physics of it as well. For example, shouldn't the fastest velocity be at the equilibrium point? Isn't the equilibrium point at t = 0 as well as its periodic multiples?
 
  • #6
diazona
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Yes, exactly. (Actually: the equilibrium point is not necessarily at t=0. It depends on how you choose your time coordinates. But it is always periodic; the oscillator will repeatedly pass through equilibrium at regular intervals.)
 
  • #7
s3a
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Ok since since we cannot assume that t = 0 when the the object ossilating is at its equilibrium point, that is why we cannot say that the maximum speed is at t = 0 and it is also why sin(pi/4) is not -1 or 1, right?
 
  • #8
diazona
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Ok since since we cannot assume that t = 0 when the the object ossilating is at its equilibrium point, that is why we cannot say that the maximum speed is at t = 0 and it is also why sin(pi/4) is not -1 or 1, right?
Right. If you like, you could figure out at what times the oscillator does pass through equilibrium for this particular problem.
 
  • #9
s3a
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I get a negative time when attempting to do it. Could you point out my mistake(s) please?
 

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  • #10
diazona
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Remember, the oscillator will pass through equilibrium at regular intervals. And unless you set t=0 to be the time that it starts moving, that applies even before t=0, at negative times. So the fact that you got a negative time doesn't necessarily mean that there's something wrong.

However, think about this: since the oscillator passes through equilibrium repeatedly, and will continue to do so over and over again, you should get not just one answer, but an infinite sequence of answers. When you invert the sine function, don't just plug arcsin into your calculator. Ask yourself "what values of [itex]\pi t + \pi/4[/itex] will make this equation true?" Then use that to solve for t.

Perhaps a more serious problem: why did you set v=1?
 

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