Finding amplitude of SHM with a theta vs time graph

In summary, the amplitude of the pendulum's motion is 0.16 meters, based on the given angular frequency of 1.26rad/s and the equation for finding amplitude using the maximum angle and string length. The problem can be solved by converting the maximum angle to radians and multiplying it by the length of the string.
  • #1
tdoyzz
8
0
1. What is the amplitude of the pendulum's motion, in meters? (see attached graph)
Given: angular frequency=1.26rad/s



2. T=2pi(L/g)^.5; a=-w^2(x); s(t)=Acos(wt+phase); theta=s/L



3. I almost don't even know how to start.. The confusing part I believe is the graph. The max angle says its 1.. so I solve for L and got 6.19m, with the period being 5s. But I still keep getting wrong answers. Please help. Thanks
 

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  • #2
tdoyzz said:
The max angle says its 1.. so I solve for L and got 6.19m,
That looks about right to me (possible rounding/precision issues though). :approve:
with the period being 5s.
Also seems reasonable. :approve:
But I still keep getting wrong answers. Please help. Thanks
The problem statement is asking for the amplitude of oscillation, not the length of the string. But since you know the length of the string (from above), and you know the max θ, how do you find the max amplitude?
 
  • #3
Would the amplitude equal the max arc length?

A = s max = theta max/L = .16

This is wrong as well though..
 
  • #4
tdoyzz said:
Would the amplitude equal the max arc length?

A = s max = theta max/L = .16

This is wrong as well though..
'Couple of things not quite right. You're dividing when you should be multiplying, and secondly, the arc length equation,
smax = θmaxL
(where smax is the arc length) only works if θmax is expressed in radians. :wink:
 
Last edited:
  • #5
Ah of course! Its always the simple things you miss.. Thanks
 

Related to Finding amplitude of SHM with a theta vs time graph

1. How do I determine the amplitude of a simple harmonic motion (SHM) from a theta vs time graph?

To find the amplitude of SHM from a theta vs time graph, you need to measure the maximum displacement of the object from its equilibrium position. This is represented by the vertical distance between the highest and lowest points on the graph. The amplitude is equal to half of this distance.

2. Can I calculate the amplitude if I only have a theta vs time graph and not the equation of motion?

Yes, you can still determine the amplitude from a theta vs time graph without the equation of motion. As mentioned in the previous answer, the amplitude is equal to half of the vertical distance between the highest and lowest points on the graph.

3. What is the unit of measurement for amplitude in SHM?

The unit of measurement for amplitude in SHM is the same as the unit of measurement for displacement, which is usually meters (m) in the SI system.

4. Does the amplitude of SHM affect the period of the motion?

Yes, the amplitude of SHM does affect the period of the motion. A larger amplitude results in a longer period, while a smaller amplitude results in a shorter period. This is because the period is directly proportional to the square root of the amplitude.

5. How do I interpret the amplitude in the context of SHM?

The amplitude in SHM represents the maximum displacement of the object from its equilibrium position. It indicates the maximum distance that the object will travel in either direction from its equilibrium point during one complete cycle of motion.

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