Simple harmonic motion problem

In summary, the answer to 3 (d) is found by calculating the gradient of the position versus time function, and differentiating that cos() function.
  • #1
tahmidbro
15
1
Homework Statement
Simple harmonic motion problem:
How is the answer to 3 (d) is found?
Relevant Equations
N/A
How is the answer to 3 (d) is found?
 

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  • #2
tahmidbro said:
Homework Statement:: Simple harmonic motion problem:
How is the answer to 3 (d) is found?
Relevant Equations:: N/A

How is the answer to 3 (d) is found?
Welcome to PF.

Are you familiar with how to find velocity from a position versus time function?
 
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  • #3
Yes. I have drawn a tangent with the curve at 'z' and calculated the gradient.
Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.
But the answer is 6m/s. Will you please tell me how?
 
  • #4
tahmidbro said:
Yes. I have drawn a tangent with the curve at 'z' and calculated the gradient.
Gradient = ( 3+3 )/(7-5) = 3 cm/s = 0.03m/s.
But the answer is 6m/s. Will you please tell me how?
I have used a derivative to calculate the velocity at that point, and also get a much smaller number than the answer key. If the amplitude is 3cm and the period is 8 seconds, the peak velocity is much less than the 6m/s that is listed as the answer.

Do you know how to do this calculation with a derivative? Also, can you check that answer with the instructor or a teaching assistant?
 
  • #5
Will you please share your calculation here? are you using v = -Aw sin(wt) ? ( this equation is not in the chapter of the exercise )
well, I am studying by myself. I do not have any instructor or a teaching assistant.
 
  • #6
tahmidbro said:
Will you please share your calculation here? are you using v = -Aw sin(wt) ? ( this equation is not in the chapter of the exercise )
Yes, good. Differentiating that cos() function does give that derivative.
 
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  • #7
w = 2pi/8 =0.785
A = 3/100 metre . For maximum velocity at 'z', sinwt = 1
v= Aw = 0.785 x 3/100 = 0.0235 m/s
But the answer from the book is 6 m/s. How do I get that?
 
  • #8
As I said, it looks like the answer provided is wrong. And not by a little bit!
 
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  • #9
berkeman said:
As I said, it looks like the answer provided is wrong. And not by a little bit!
Okay , Thanks for the help!
 
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Related to Simple harmonic motion problem

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction of the displacement. This results in a sinusoidal or oscillatory motion.

2. How do you calculate the period of a simple harmonic motion?

The period of a simple harmonic motion can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant.

3. What is the relationship between frequency and period in simple harmonic motion?

The frequency of a simple harmonic motion is the number of oscillations per unit time, while the period is the time it takes for one complete oscillation. The relationship between frequency and period is inverse, meaning that as the frequency increases, the period decreases, and vice versa.

4. How does amplitude affect the motion in simple harmonic motion?

The amplitude of a simple harmonic motion is the maximum displacement from equilibrium. The amplitude affects the motion by determining the maximum potential energy and the maximum velocity of the object. A larger amplitude will result in a greater maximum potential energy and a higher maximum velocity.

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

In simple harmonic motion, there is no external force acting on the system, resulting in a constant amplitude and frequency. In damped harmonic motion, there is an external force, such as friction, that causes the amplitude and frequency to decrease over time, eventually leading to the system coming to rest at equilibrium.

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