Why period is half ? Spring problem

  • Thread starter Thread starter k31453
  • Start date Start date
  • Tags Tags
    Period Spring
AI Thread Summary
The discussion centers on a mass sliding on a frictionless surface that strikes a spring, raising questions about the period of oscillation and the duration of contact with the spring. The period of oscillation is derived from the spring constant and mass, leading to the conclusion that the mass is in contact with the spring for half of the oscillation period. This is because a full cycle involves the mass moving through maximum compression and returning to its original position, while contact occurs only during the compression phase. The key takeaway is that the mass loses contact with the spring when the spring force reaches zero, indicating it only engages with the spring for half the period. Understanding these dynamics clarifies why the period is effectively halved in this scenario.
k31453
Messages
58
Reaction score
0

Homework Statement


A mass m slides along a frictionless horizontal surface at speed v initial. It strikes a spring of constant k attached to a rigid wall. After a completely elastic encounter with the spring, the mass heads back in the direction it came from.

(a)In terms of k, m, and v initial, determine how long the mass is in contact with the spring.

(b)In terms of k, m, and v initial, determine the maximum compression of the spring.


Homework Equations



The one thing i don't know is why period = 1/2 ??


The Attempt at a Solution



I know horizontal force will be only the force exert by spring.
I know the algebra below:
f= 1/2π * √(k/m)

Period = P = 1/f ergo
t = P/2
= 1/2* 1/f
= 1/2 * 2π √(m/k)
= π √(m/k)
 
Last edited:
Physics news on Phys.org
k31453 said:
The one thing i don't know is why period = 1/2 ??

Think about the difference between this scenario and the "usual" spring oscillator where the mass is fixed to the end of the spring. What constitutes a full cycle of the oscillation?
 
I think you mean that the mass will be in contact with the spring for half the period of oscillation (if the mass was attached to the spring).
 
Yes BASIC But how ?

And I know when max displacement the T is going to be 1.
 
k31453 said:
Yes BASIC But how ?

And I know when max displacement the T is going to be 1.

You should listen to what gneill said. Suppose you started at zero compression or stretch (as is the case here) and max inward speed. If the mass had been attached, what would *happen* during a full period of oscillation?

How much of that happens here? Hint: the mass loses contact with the spring when the spring force goes to zero.
 
half period

Assume the mass is attached to the spring - a necessary condition for SHM to occur.
At position 1 the spring is relaxed.
At position 2 the mass is at its rightmost extreme position and the spring is at max compression.
At position 3 the spring is back to its relaxed state.
At position 4 the mass is at its leftmost extreme position and the spring is at max extension.
At position 5 the spring is again relaxed.
 

Attachments

  • half period.jpg
    half period.jpg
    11.9 KB · Views: 403
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top