Simple Harmonic Motion Given Amplitude and Frequency

In summary, to find the maximum magnitude of the velocity for a cheerleader waving her pom-pom in SHM with an amplitude of 17.3 cm and a frequency of 0.830 Hz, one can use the equation v = -ωA, where ω is equal to 2π/T with T being the period of the motion. Using this equation, the maximum magnitude of the velocity is found to be 0.901 m/s. Additionally, to find the speed when the pom-pom's coordinate is x= 9.40 cm, one can use the equation v = ωx, where ω is the angular frequency and x is the displacement. Using this equation, the speed is found to be
  • #1
Vanessa Avila
94
1

Homework Statement


A cheerleader waves her pom-pom in SHM with an amplitude of 17.3 cm and a frequency of 0.830 Hz .
Find the maximum magnitude of the velocity.

Homework Equations


v = -w Asin(ωt+Φ) = -wx
or
Conservation of Energy:
1/2kx2 + 1/2mv2 = 1/2kA2

The Attempt at a Solution


I tried v = -ω(x) using 0.173 as x
v = -5.21(0.173) = -0.0901 > wrong

I got the ω by solving for period T (1.205s), i then made this equal to 2π/ω to solve for omega.

I have yet to try the conservation of energy. But what would be the x in this case?
 
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  • #2
I just realized that i had an extra 0 there for -0.0901, should have been 0.901!
 
  • #3
Vanessa Avila said:

Homework Statement


A cheerleader waves her pom-pom in SHM with an amplitude of 17.3 cm and a frequency of 0.830 Hz .
Find the maximum magnitude of the velocity.

Homework Equations


v = -w Asin(ωt+Φ) = -wx
or
Conservation of Energy:
1/2kx2 + 1/2mv2 = 1/2kA2

The Attempt at a Solution


I tried v = -ω(x) using 0.173 as x
v = -5.21(0.173) = -0.0901 > wrong
There are to many 0-s. And the magnitude is not negative!
 
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  • #4
ehild said:
There are to many 0-s. And the magnitude is not negative!
Thanks! :)
 
  • #5
ehild said:
There are to many 0-s. And the magnitude is not negative!
I do have another question actually. How would i solve this?
"Find the speed when the pom-pom's coordinate is x= 9.40 cm ."

I tried to use the equation v = ωx and got 0. 48974, but i got the wrong answer
v = (5.21rad/s)(0.094m) = 0.4897
 
  • #6
Vanessa Avila said:
I do have another question actually. How would i solve this?
"Find the speed when the pom-pom's coordinate is x= 9.40 cm ."

I tried to use the equation v = ωx and got 0. 48974, but i got the wrong answer
v = (5.21rad/s)(0.094m) = 0.4897

The velocity is the time derivative of the displacement. If x=Acos(ωt) what is v=dx/dt?
V=ωA is the relation between maximum displacement and maximum speed.
 
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1. What is Simple Harmonic Motion (SHM)?

SHM is a type of periodic motion in which an object moves back and forth along a straight line with a constant amplitude and frequency. It is a common type of motion observed in many physical systems, such as a pendulum or a mass-spring system.

2. What is the relationship between amplitude and frequency in SHM?

The amplitude of an object in SHM is directly proportional to its frequency. This means that as the frequency increases, the amplitude also increases. Conversely, as the frequency decreases, the amplitude decreases as well.

3. How are amplitude and frequency related to the period of SHM?

The period of SHM is the time it takes for one complete cycle of motion. It is inversely proportional to the frequency and directly proportional to the amplitude. This means that as the frequency increases, the period decreases, and as the amplitude increases, the period also increases.

4. Can the amplitude and frequency of SHM be changed?

Yes, the amplitude and frequency of SHM can be changed by altering the physical characteristics of the system, such as the length of a pendulum or the stiffness of a spring. Additionally, the amplitude and frequency can also be changed by applying external forces to the system.

5. What is the equation for calculating the period of SHM?

The period of SHM 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 or the restoring force constant of the system.

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