What Speed Ensures Maximum Amplitude for a Mass on a Spring System?

  • Thread starter Thread starter cutegirl1980
  • Start date Start date
  • Tags Tags
    Amplitude
AI Thread Summary
To achieve a maximum amplitude of 4 cm for a 2 kg mass on a spring with a spring constant of 13 N/m, the necessary speed is calculated to be 26.1 cm/s. The amplitude of the road is 2 cm, and the mass's amplitude must be double that, hence 4 cm. The angular frequency is determined using the formula ω = √(k/m), resulting in approximately 2.55 rad/s. The time taken to reach the maximum amplitude is found to be 0.613 seconds. This analysis emphasizes the importance of using the correct equations for simple harmonic motion to solve for unknown variables effectively.
cutegirl1980
Messages
3
Reaction score
0
box of mass 2kg on spring with k=13 N/m, they are on a wheel. The road has an amplitude δ = 2 cm as shown above. What speed is necessary for the
mass to have a maximum amplitude x = 2δ . the picture looks something like this
__
|__| m=2kg
\
/ k=13 N/m
\
/--\__________________
\--/ 16.0 cm

^ the thing the bottom of the spring is connected to looks like a wheel. the distance is 16 cm. can anyone help me, very confusing question.
 
Physics news on Phys.org
This is a simple harmonic motion problem with a forcing function.

Write the equation for SHO without damping and with a forcing function.

See http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html for some basic equations.

You know that the amplitude of the mass on the spring has to be twice the amplitude of the forcing function, and you have to find the frequency of the forcing function that will give you that amplitude.

Remember how to find the frequency of a mass (m) on a spring with constant (k).
 


To solve this problem, we can use the equation for the maximum amplitude of a mass-spring system:
x_max = A*sin(ωt)
Where A is the amplitude, ω is the angular frequency, and t is time.
We know that the amplitude of the road is δ = 2 cm, so the maximum amplitude of the mass on the spring would be x_max = 2δ = 4 cm.
We can also calculate the angular frequency ω using the spring constant k and mass m:
ω = √(k/m)
Plugging in the values given in the problem, we get ω = √(13 N/m / 2 kg) = √(6.5) rad/s.
Now, we can set x_max = A*sin(ωt) equal to 4 cm and solve for t:
4 cm = A*sin(√(6.5) rad/s * t)
Solving for t, we get t = 0.613 seconds.
Since we know that the distance between the bottom of the spring and the wheel is 16 cm, we can calculate the speed needed for the mass to reach this distance in 0.613 seconds:
v = d/t = 16 cm / 0.613 s = 26.1 cm/s.
Therefore, the speed necessary for the mass to have a maximum amplitude of 4 cm is 26.1 cm/s.
I hope this helps you solve the problem. Remember to always carefully read the given information and use the appropriate equations to solve for the unknown variables.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top