Simple Harmonic Motion question.

In summary, the two pendulums will once again be in step after 6.86 seconds and they will have gone through 5.5 oscillations each.
  • #1
gabloammar
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Homework Statement


Two simple pendulums of length 0.40 m and 0.60 m are set off oscillating in step. Calculate (a) after what further time the two pendulums will once again be in step, (b) the number of oscillations made by each pendulum during this time. (Assume g = 10 ms-2

Homework Equations



T = 2[itex]\pi\sqrt{l/g}[/itex]

The Attempt at a Solution



I get the two different values for T, but I'm stuck after that. I've got the solution with me but I need someone to explain getting the answer to me step by step I suppose. If anyone needs the solution I'd be happy to type it all out. Thanks!
 
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  • #2
Yes, what is your actual attempt at the solution? We can show you the next step with it.
 
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  • #3
No see MY attempt at the solution was finding T first, and after that I'm stuck. What I said above is that I've got the solution written with me in the textbook and that if you need the whole solution to tell me what's happening then I'll gladly type it all out or scan it in a while and post it here and maybe then you can help me. However it suits you :)
 
  • #4
what does it mean by 'set off oscillating in step'? Are these coupled or separate pendulums? If they are separate, then it seems like a sort of trick question...
 
  • #5
gabloammar said:

Homework Statement


Two simple pendulums of length 0.40 m and 0.60 m are set off oscillating in step. Calculate (a) after what further time the two pendulums will once again be in step, (b) the number of oscillations made by each pendulum during this time. (Assume g = 10 ms-2


Homework Equations



T = 2[itex]\pi\sqrt{l/g}[/itex]

The Attempt at a Solution



I get the two different values for T, but I'm stuck after that. I've got the solution with me but I need someone to explain getting the answer to me step by step I suppose. If anyone needs the solution I'd be happy to type it all out. Thanks!

Sorry, can't do the step by step, it's against board rules. But I'll give you a big hint:

The number of cycles an oscillator has gone through after time [itex]t[/itex] can be represented by [itex]\frac{t}{T}[/itex], where [itex]T[/itex] is the period. If this number is an integer, it means it's undergone that many complete oscillations within that time. If the number is not an integer, it means that it is in the "middle" of an oscillation at the time.

To be in sync, two oscillators which have been synchronised at the start need to have a cycle number difference which is a whole number. The smallest integer is zero (this occurs at the start). The next integer is one.

Can you work out the solution now?
 
  • #6
Curious3141 said:
Sorry, can't do the step by step, it's against board rules. But I'll give you a big hint:

The number of cycles an oscillator has gone through after time [itex]t[/itex] can be represented by [itex]\frac{t}{T}[/itex], where [itex]T[/itex] is the period. If this number is an integer, it means it's undergone that many complete oscillations within that time. If the number is not an integer, it means that it is in the "middle" of an oscillation at the time.

To be in sync, two oscillators which have been synchronised at the start need to have a cycle number difference which is a whole number. The smallest integer is zero (this occurs at the start). The next integer is one.

Can you work out the solution now?

[Sorry just read the rules again, didn't read 'em that well before I suppose. But I understand why you can't help me step by step and that's fine since it's the rules.]

Secondly, YES! See I've got the solution and everything but I didn't understand why we needed to take the second oscillation as (n-1), I understand the rest of it. Was just stuck at this part. Thank you SO much!
 
  • #7
what answer did you arrive at?
 
  • #8
For (a), I got 6.86 s. And for (b), the shorter pendulum makes 5.5 oscillations and the longer pendulum makes 4.5 oscillations :)
 
  • #9
And this is the same as the answer given in the textbook? Strictly speaking, the two are not exactly in step again. But they are close to being in step.
 
  • #10
Yep, 'cause I got to round 'em off to two significant figures.
 
  • #11
But what did you round off to two significant figures? the number of oscillations? So the number of oscillations shouldn't be 5.5 and 4.5 ? But then they would not be in sync.

I guess I'm trying to argue with the textbook's answer. And you've probably moved on from this question a long time ago. So don't worry about it.
 

FAQ: Simple Harmonic Motion question.

What is simple harmonic motion?

Simple harmonic motion refers to the back and forth motion of an object around a stable equilibrium point, where the object's acceleration is directly proportional to its displacement from the equilibrium point and directed towards it. This type of motion is also characterized by a sinusoidal or wave-like pattern.

What are the factors that affect simple harmonic motion?

The main factors that affect simple harmonic motion are the mass of the object, the spring constant of the system, and the amplitude or maximum displacement of the object from the equilibrium point. Any changes in these factors can result in a change in the frequency and period of the motion.

What is the equation for simple harmonic motion?

The mathematical equation for simple harmonic motion is x(t) = A cos(ωt + φ), where x is the displacement of the object from the equilibrium point at time t, A is the amplitude, ω is the angular frequency, and φ is the phase constant. This equation can also be written in terms of velocity and acceleration by taking the first and second derivatives, respectively.

What is the difference between simple harmonic motion and damped harmonic motion?

In simple harmonic motion, the object continues to oscillate back and forth without any external forces acting on it. In contrast, damped harmonic motion involves the presence of external forces, such as friction or air resistance, which causes the oscillations to gradually decrease in amplitude and eventually come to a stop.

What are some real-life examples of simple harmonic motion?

Some common examples of simple harmonic motion include the motion of a pendulum, the back and forth movement of a spring, and the swinging of a playground swing. Other examples include the vibration of guitar strings, the motion of a mass attached to a vertical spring, and the movement of a mass on a horizontal spring with a constant force acting on it.

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