How Do You Calculate Kinetic Energy and Frequency in Harmonic Motion Problems?

In summary: Now, while the jumper is in the air, and before they hit the ground,PE + KE = 1/2 k*x2But that is a constant (call it C), sod(PE + KE) = dC = 0dPE + dKE = 0dPE = -dKENow we can talk about the energy at 2 sec, but not before.2. I will not provide a summary for this conversation as it is not relevant to the summary requested.
  • #1
nns91
301
1

Homework Statement



1. The St. Louis Arch has a height of 192m. Suppose a stunt woman of mass 60kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. Find her kinetic energy K after 2 sec of the flight. Assume that the elastic band obeys Hooke's law, and neglect its length when relaxed

2. A 0.12kg block is suspended from a spring. When a small stone of mass 30g is placed on the block, the spring stretches an additional 5cm. With the stone on the block, the spring oscillates with an amplitude of 12cm. (a) What is the frequency of the motion ? (b) How long does the block take to travel from its lowest point to its highest point ? (c) What is the net force of the stone when it is at a point of maximum upward displacement ?

Homework Equations



w= sqrt(k/m)

The Attempt at a Solution



1. I use ky=mg to calculate k and get k=60*9.81/192=3.066 N/m
Then I calculate w=sqrt(k/m)=0.226 rad/s

Then I use v=-A*w*sin(wt) to calculate v at t=2 and get -19m/s and substitute into K=1/2 mv^2 and got 10.8kJ. However, I am wrong. Can you guy suggest me a way to fix my solution ??

2. I used the same formula as number 1: ky=mg to get k and then w=sqrt(k/m)=9.04 rad/s. Finally I use f=w/2pi to get f=1.44Hz.

For part b, T=1/f=0.694 sec and for part c, F=mg+kx= 119 N

I am wrong again.Any suggestion ??
 
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  • #2
nns91 said:

Homework Statement



1. The St. Louis Arch has a height of 192m. Suppose a stunt woman of mass 60kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. Find her kinetic energy K after 2 sec of the flight. Assume that the elastic band obeys Hooke's law, and neglect its length when relaxed

2. A 0.12kg block is suspended from a spring. When a small stone of mass 30g is placed on the block, the spring stretches an additional 5cm. With the stone on the block, the spring oscillates with an amplitude of 12cm. (a) What is the frequency of the motion ? (b) How long does the block take to travel from its lowest point to its highest point ? (c) What is the net force of the stone when it is at a point of maximum upward displacement ?

Homework Equations



w= sqrt(k/m)

The Attempt at a Solution



1. I use ky=mg to calculate k and get k=60*9.81/192=3.066 N/m

Stop right there. When the jumper reaches the ground with V = 0, and 0 kinetic energy, she is not in static equilibrium. Hence you need to examine the energy relationship.

At the top the total energy is m*g*h, so neglecting the relaxed length, and the practicality of trying to envision it, that means that the PE at the top must equal the PE of the elastic at the bottom. So ...

m*g*h = 1/2* k*x2

Or k = 2mg/h
 
  • #3


Your approach to the first problem looks correct, but it seems like you may have made a mistake in your calculations. I would recommend double-checking your values for k and w, and then re-doing the calculations to find the kinetic energy. It may also be helpful to draw a diagram and label all the variables to make sure you are using the correct values.

For the second problem, your approach using Hooke's law and the equations for frequency and period are correct. However, for part c, the net force at the maximum upward displacement would be the sum of the force due to gravity and the force from the spring, which would be equal to the weight of the block plus the weight of the stone. So the net force would be 0.12*9.81 + 0.03*9.81 = 1.17 N. Once again, it may be helpful to draw a diagram to visualize the forces at different points in the motion.
 

1. What is spring harmonic motion?

Spring harmonic motion is a type of vibrational motion in which a mass attached to the end of a spring oscillates back and forth around an equilibrium position. The motion is periodic and can be described by a sinusoidal function.

2. What factors affect the frequency of spring harmonic motion?

The frequency of spring harmonic motion is primarily affected by the mass of the object attached to the spring and the stiffness of the spring. A heavier mass or a stiffer spring will result in a higher frequency of oscillation.

3. How is spring constant related to spring harmonic motion?

The spring constant, also known as the force constant, is a measure of the stiffness of a spring. It is directly proportional to the frequency of spring harmonic motion, meaning that as the spring constant increases, the frequency of oscillation also increases.

4. How can spring harmonic motion be used in real-world applications?

Spring harmonic motion has many practical applications, such as in watches, suspension systems for vehicles, and musical instruments. It can also be used in scientific experiments to study the properties of materials.

5. What is the formula for calculating the period of spring harmonic motion?

The period of spring harmonic motion can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the mass in kilograms, and k is the spring constant in newtons per meter.

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