Block Oscillating on Spring

In summary, the problem involves an oscillating block on a spring in a fluid with a resistive force. To find the period of the motion, we can use the formula T = 2π√(m/k) and solve for the length l by equating it to √(g/l). The period is approximately 1.102 seconds. For part b, we can use the formula for damped frequency f' = 1/(2π)√(k/m - b^2/(4m^2)) and solve for the fractional decrease in amplitude per cycle. For part c, we can use the equation x(t) = x_0 + Ae^(-Γt)cos(ω't
  • #1
gmmstr827
86
1

Homework Statement



A 0.835-kg block oscillates on the end of a spring whose spring constant is k = 41.0 N/m. The mass moves in a fluid which offers a resistive force F = -bv, where b = 0.662 Ns/m.
a) What is the period of the motion?
b) What is the fractional decrease in amplitude per cycle?
c) Write the displacement as a function of time if at t = 0, x = 0, and t = 1.00 s, x = 0.120m.

m = 0.835 kg
k = 41.0 N/m
b = 0.662 Ns/m
g = -9.80 m/s^2

Homework Equations



F = -bv
T (period) = 2*π*√(m/k)
ε = √(k/m) = √(g/l)
l = length

The Attempt at a Solution



a) Since l is not given, I found it, since ε = √(k/m) and ε = √(g/l), therefore √(k/m) = √(g/l)
and l = .1996 m after plugging in the givens and solving for l.
Now, the rest I'm not as sure of:
F = -b*v = -(0.662 Ns/m)(-9.8 m/s^2) ≈ 6.49 N/s
So, using that value in place of g:
T (period) = 2π√(.1996m/6.49N/s) ≈ 1.102 s

^^^ Is that correct? How would I solve for b and c?
 
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  • #2
first find equilibrium position. note that resistive force depends on velocity of block. it is damped SHM. use the proper equations, if you don't know, refer a theory book.
 
  • #3
I've revamped my part a using more appropriate formulas.
In order to find the equilibrium first, I would need to use this formula:
x(t) = x_0 + Ae^(-Γt)cos(ω't + φ) where x_0 = 0; ω' = √[k/m - b^2/(2m^2)]; Γ = b/(2m); Φ = arctan{-v(0)/[ε*x(0)]}
The problem is, A is not given and I do not have a formula to solve for A.
However, I believe I was able to find the period despite this.
f' (damped frequency) = ε'/(2π) = [1/(2π)]*√[(k/m)-(b^2/(4m^2))]
f' = [1/(2π)]*√[(41.0 N/m / 0.834 kg)-((0.662 N/m)^2/(4*(0.835 kg)^2))] ≈ 1.114 Hz
T' = 1/f' ≈ 1/1.114 Hz ≈ 0.898 s

So, would one Period be approximately 0.898 seconds?

For part b, I was going to use the formula x(t) = x_0 + Ae^(Γt)cos(ω't + φ), but I do not have a way to find amplitude, and would end up with multiple unsolved variables.

It seems like part c would include the same equation? Would I be solving for x_0 in part c? It would seem that it would go as follows:

x(t) = x_0 + Ae^(-Γt)cos(ω't + φ) where ω' = √[k/m - b^2/(2m^2)]; Γ = b/(2m); Φ = arctan{-v(0)/[ε*x(0)]}

Where t=0, x=0
0 = x_0 + Acos(- φ)
x_0 = -Acos(-φ)

Where t=1.00s, x=.120m
.120 m = x_0 + Ae^(-0.662/(2(0.835 m))cos(√[41.0 N/m / 0.835 kg - 0.662^2/(2(0.835 kg)^2)] + φ)
x_0 = -0.120 m - Acos(√(48.789)+φ)/e^(0.3964)

Am I on the right track here? Any hints/solution styles would be greatly appreciated.
 
  • #4
Does anyone know if this looks right so far..? The biggest tips I need now are what to do with part b and if I should instead solve for x instead of x_0 in part c, making x = 0 instead of x_0 = 0.
 
  • #5


b) To find the fractional decrease in amplitude per cycle, we can use the equation for damping ratio, ζ = b/2√(mk), where m is the mass and k is the spring constant. Plugging in the given values, we get ζ = 0.662/(2√(0.835*41.0)) ≈ 0.057. This means that the amplitude will decrease by approximately 5.7% per cycle.

c) To write the displacement as a function of time, we can use the equation x(t) = A*e^(-ζωt)*cos(ωt + φ), where A is the amplitude, ζ is the damping ratio, ω is the angular frequency, and φ is the phase angle. We can find ω by using the equation ω = √(k/m) = √(41.0/0.835) ≈ 7.27 rad/s. The phase angle φ can be found by using the given information at t = 1.00 s, x = 0.120 m. Substituting these values into the equation, we get x(t) = 0.120*e^(-0.057*7.27*1.00)*cos(7.27*1.00 + φ). Solving for φ, we get φ ≈ -0.024 rad. Therefore, the displacement as a function of time is x(t) = 0.120*e^(-0.057*7.27*t)*cos(7.27*t - 0.024).
 

1. What is a block oscillating on a spring?

A block oscillating on a spring is a simple mechanical system in which a mass (the block) is attached to a spring and allowed to move back and forth. The spring provides a restoring force that causes the block to oscillate, or move back and forth repeatedly.

2. How does the spring affect the motion of the block?

The spring provides a restoring force that is directly proportional to the displacement of the block from its equilibrium position. This means that the further the block is stretched or compressed from its resting point, the greater the force that the spring exerts on it, causing it to accelerate and oscillate.

3. What factors affect the frequency of oscillation?

The frequency of oscillation of a block on a spring is affected by three main factors: the mass of the block, the stiffness of the spring (measured by its spring constant), and the amplitude of oscillation (how far the block moves from its equilibrium position). Generally, a larger mass or a stiffer spring will result in a lower frequency of oscillation, while a larger amplitude will result in a higher frequency.

4. How does damping affect the oscillation?

Damping, or the gradual loss of energy from the system, can affect the amplitude and frequency of oscillation. In a block on a spring system, damping can be caused by friction between the block and the surface it is resting on, or by air resistance. Damping can cause the amplitude of oscillation to decrease over time, and may also affect the frequency of oscillation.

5. What real-life applications does a block oscillating on a spring have?

Block oscillating on springs are used in many real-life applications, including shock absorbers in vehicles, earthquake-resistant buildings, and tuning forks in musical instruments. They are also used in scientific experiments to study simple harmonic motion and the effects of damping.

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