Oscillating mass on a spring - non-constant mass

Click For Summary
SUMMARY

The discussion revolves around calculating the rate of change of the oscillation period of a bucket with a leak, containing water, suspended from a spring. The bucket has a mass of 2.10 kg and initially contains 13.0 kg of water, with a spring constant of 130 N/m and an oscillation amplitude of 3.00 cm. The water leaks at a rate of 2.00 g/s. To find the changing period, participants suggest using the formula T=2π√(Σm/k) and applying the Chain Rule to derive the period as a function of time, particularly when the bucket is half full.

PREREQUISITES
  • Understanding of harmonic motion and oscillation principles
  • Familiarity with the formula for the period of a spring-mass system
  • Knowledge of the Chain Rule in calculus
  • Ability to differentiate functions with respect to time
NEXT STEPS
  • Study the derivation of the period formula for spring-mass systems
  • Learn how to apply the Chain Rule in physics problems
  • Investigate the effects of mass change on oscillation frequency
  • Explore examples of non-constant mass systems in oscillatory motion
USEFUL FOR

Students in physics, particularly those studying mechanics and oscillatory systems, as well as educators looking for examples of real-world applications of calculus in physics.

Nivlac2425
Messages
53
Reaction score
0

Homework Statement


A 2.10 kg bucket containing 13.0 kg of water is hanging from a vertical ideal spring of force constant 130 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s.

When the bucket is half full, find the rate at which the period is changing with respect to time.

Homework Equations


T=2\pisqrt(Ʃm/k)

The Attempt at a Solution


I know that I need to find T as a function of t, then take the derivative wrt t and evaluate it at the time when the bucket is half full.
But I'm not sure how to set this up.. I tried:
T=2\pisqrt[(m1+m2+Δmt)/k]
where m1=2.1 kg, m2=13 kg, and Δm=0.02 kg/s
but it seems to be the wrong set-up. Anyone have any ideas?

Also, when the bucket is half full, is t=(half the mass of water)/(0.02 kg/s)=6.5/0.02=325 seconds? Is this the t that we should evaluate the derivative at? (assuming we figure it out first :-p)

Thanks for helping!
 
Physics news on Phys.org
Nivlac2425 said:

Homework Statement


A 2.10 kg bucket containing 13.0 kg of water is hanging from a vertical ideal spring of force constant 130 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s.

When the bucket is half full, find the rate at which the period is changing with respect to time.

Homework Equations


T=2\pisqrt(Ʃm/k)

The Attempt at a Solution


I know that I need to find T as a function of t, then take the derivative wrt t and evaluate it at the time when the bucket is half full.
But I'm not sure how to set this up.. I tried:
T=2\pisqrt[(m1+m2+Δmt)/k]
where m1=2.1 kg, m2=13 kg, and Δm=0.02 kg/s
but it seems to be the wrong set-up. Anyone have any ideas?

Also, when the bucket is half full, is t=(half the mass of water)/(0.02 kg/s)=6.5/0.02=325 seconds? Is this the t that we should evaluate the derivative at? (assuming we figure it out first :-p)

Thanks for helping!

You need to find T as a function of m, which you've already done. m consists of a fixed component (the bucket) and a decreasing component (the leaking water).

Then, to get an expression for the rate of change of the period with respect to time, use the Chain Rule. You don't actually have to explicitly derive an expression for m in terms of t.

Once you get the expression, it's as simple as figuring out the mass when the bucket is half full and putting this into the expression.

BTW, if you're taking the tack of actually deriving an expression for m in terms of t, then remember that the rate of change of mass is NEGATIVE 0.002 kg/s (-0.002 kg/s) - note the sign and the number of zeros.

Your method to work out the time when the bucket is half-full is correct in principle, but you have the rate of leakage too high by an order of magnitude, as mentioned above.
 
Last edited:
Curious3141 said:
You need to find T as a function of m, which you've already done. m consists of a fixed component (the bucket) and a decreasing component (the leaking water).

Then, to get an expression for the rate of change of the period with respect to time, use the Chain Rule. You don't actually have to explicitly derive an expression for m in terms of t.

Once you get the expression, it's as simple as figuring out the mass when the bucket is half full and putting this into the expression.

BTW, if you're taking the tack of actually deriving an expression for m in terms of t, then remember that the rate of change of mass is NEGATIVE 0.002 kg/s (-0.002 kg/s) - note the sign and the number of zeros.

Your method to work out the time when the bucket is half-full is correct in principle, but you have the rate of leakage too high by an order of magnitude, as mentioned above.

Thanks for pointing out what I needed! I've figured it out, and I really appreciate the help :biggrin:
 

Similar threads

Replies
17
Views
2K
Replies
4
Views
947
Replies
20
Views
2K
Replies
1
Views
1K
Replies
31
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
3
Views
2K