Shm , calculation of amplitude of spring mass system

In summary, the spring stretches 10 cm from it's free length when a mass of 2 kg is attached and released from rest. The force constant of the spring is 500 N/m and the amplitude of the vibration is 10 cm.
  • #1
Shivang kohlii
19
1

Homework Statement


In A spring mass system , the spring stretches 2 cm from its 's frelength when a force of 10 N is applied . This spring is stretched 10 cm from it's free length , when a body of mass m = 2 kg is attached to it and released from rest at time t = 0 . Find the A) force constant of spring
B) amplitude of vibration

Homework Equations


kx = mg
X is the extension from mean position
Elastic potential energy = 1/2 kx^2

The Attempt at a Solution


A) k = 10 / 0.02 = 500N/m
B ) considering the mean position as the point of extension of spring due to mass m ,
Energy at mean position ( X= 0 )= energy at maximum displacement( X= Amplitude , A )

Kinetic energy at mean position
=1/2 m×ω^2× A
= (1/2) × 2 × (k/m) × A
= 250 × A Joules

Gravitational potential energy at mean position ( taking base as the extreme position ) = mgA =
2 × 10 × A = 20A

Spring potential energy at mean position = 0

ii) kinetic energy at extreme position = gravitational potential energy at extreme position = 0
Elastic potential energy at extreme position = 1/2 × 500 × (A)^2 = 250 ×A^2

Equating the energies :
A = 1.8 but answer is 0.06 m

I am wondering if I made some mistake in the initial energy ? If yes , please point the mistake out with correction.. any help will be much appreciated ! Thanks
 
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  • #2
Shivang kohlii said:
Spring potential energy at mean position = 0
The spring is not relaxed at equilibrium.

There is a much easier way. Just find the extension at equilibrium.
 
  • #3
haruspex said:
The spring is not relaxed at equilibrium.

There is a much easier way. Just find the extension at equilibrium.
Yes I know the method through extension at equilibrium ... But I want to check out the method of COE for future purposes...
Can u please just tell me the equation that u would use and the two points that u would consider for solving through COE ?
 
  • #4
Shivang kohlii said:
Can u please just tell me the equation that u would use and the two points that u would consider for solving through COE ?
Try again for the spring PE at equilibrium.
 

1. What is a "Shm" or "simple harmonic motion" system?

A "Shm" or "simple harmonic motion" system is a type of system in physics that involves the periodic back-and-forth motion of an object due to the restoring force of a spring. It is characterized by a constant amplitude and frequency.

2. How do you calculate the amplitude of a spring mass system?

The amplitude of a spring mass system can be calculated by dividing the maximum displacement of the mass by 2. This value represents the distance from the equilibrium position to the maximum displacement of the mass in either direction.

3. What factors affect the amplitude of a spring mass system?

The amplitude of a spring mass system is affected by the mass of the object, the stiffness of the spring, and the initial conditions (position and velocity) of the object. A heavier mass and a stiffer spring will result in a larger amplitude, while changing the initial conditions can alter the amplitude as well.

4. How does changing the amplitude of a spring mass system affect its frequency?

Changing the amplitude of a spring mass system does not affect its frequency. The frequency is solely determined by the mass and stiffness of the system, and remains constant regardless of the amplitude.

5. Can the amplitude of a spring mass system be negative?

Yes, the amplitude of a spring mass system can be negative. This indicates that the object is moving in the opposite direction of its initial displacement. However, the magnitude of the amplitude remains the same regardless of its direction.

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