Calculate the total energy of the mass spring system

Click For Summary
SUMMARY

The total energy of a mass-spring system with a 12 kg mass and a spring constant of 320 N/m, oscillating with an amplitude of 10 cm, is calculated to be 1.6 Joules using the formula ET=(0.5)(k)(A)^2. The maximum speed of the mass is determined to be 0.52 m/s, derived from the energy conservation equation. Additionally, when the mass is 5.0 cm from the equilibrium point, its speed is calculated to be 0.45 m/s, confirming the application of energy principles in simple harmonic motion.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with Hooke's Law (Fx = -kx)
  • Knowledge of energy conservation in mechanical systems
  • Ability to manipulate equations involving kinetic and potential energy
NEXT STEPS
  • Study the derivation of the maximum speed in simple harmonic motion
  • Learn about energy transformations in oscillatory systems
  • Explore the effects of varying mass and spring constants on oscillation
  • Investigate the damping effects in real-world mass-spring systems
USEFUL FOR

Students in physics, mechanical engineers, and anyone studying dynamics and oscillatory motion will benefit from this discussion.

AClass
Messages
23
Reaction score
0

Homework Statement


A 12Kg mass udnergoes simple harmonic motion with an amplitude of 10cm. If the force constant for the spring is 320 N/m

a) Calculate the total energy of the mass spring system.
b) Calculate the maximum speed of the mass
c) Calculate the speed of the mass when it is 5.0cm from the equilibrium point.


Homework Equations



Fx= -kx
a=kx/m
ET=(0.5)mv^2 +(0.5)kx^2
Et=(0.5)(m)(A)^2

The Attempt at a Solution



a)

ET=(0.5)(320 N/m)(0.10m)^2
ET= 1.6J

b)

1.6J=(0.5)(12kg)(V)^2 + (0.5)(320N/m)(0)
v=0.52m/s

c)

1.6J=(0.5)(12kg)(V)^2 +(0.5)(320N/m)(0.05m)^2
v=0.45m/s

Could someone verify this for me? Thanks
 
Physics news on Phys.org
Those look correct to me (the method, I did not check the numbers).
 
1.a) E=1/2KA^2=1/2(320N/m)(0.1m)^2=1.6J b) V(max)=wA
 

Similar threads

Replies
17
Views
2K
Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
24
Views
4K