Oscillating mass on a spring - non-constant mass

In summary, the problem involves a bucket of water hanging from a vertical ideal spring and oscillating up and down. When the bucket is half full, the rate at which the period is changing with respect to time needs to be found. Using the Chain Rule, an expression for the rate of change of the period with respect to time can be derived. The rate of change of mass, which is negative 0.002 kg/s, needs to be taken into account. The time when the bucket is half full is calculated to be 325 seconds.
  • #1
Nivlac2425
54
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[itex]\pi[/itex]sqrt(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[itex]\pi[/itex]sqrt[(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 :tongue2:)

Thanks for helping!
 
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  • #2
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[itex]\pi[/itex]sqrt(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[itex]\pi[/itex]sqrt[(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 :tongue2:)

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:
  • #3
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:
 

What is an oscillating mass on a spring?

An oscillating mass on a spring is a physical system where a mass is attached to a spring and allowed to move back and forth repeatedly due to the restoring force of the spring. This type of system is often studied in physics to understand the principles of simple harmonic motion.

How does the mass affect the oscillation of the spring?

The mass attached to the spring affects the oscillation by determining the period and frequency of the motion. A larger mass will result in a longer period and lower frequency, while a smaller mass will result in a shorter period and higher frequency. However, the amplitude of the oscillation remains the same regardless of the mass.

Does the mass have to be constant for the oscillation to occur?

No, the mass does not have to be constant for the oscillation to occur. As long as the mass is within the range that the spring can support, the oscillation will still occur. However, changing the mass can affect the properties of the oscillation, such as the period and frequency.

How does the spring constant affect the oscillation?

The spring constant, also known as the stiffness of the spring, affects the oscillation by determining the strength of the restoring force. A higher spring constant will result in a stronger restoring force, leading to a shorter period and higher frequency of oscillation.

What factors can affect the oscillation of the mass on a spring?

Aside from the mass and spring constant, other factors that can affect the oscillation include the initial displacement of the mass, external forces such as friction or air resistance, and the properties of the spring, such as its length and shape.

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