Block Oscillating on Spring

  • Thread starter gmmstr827
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
supratim1
Gold Member
279
1
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
86
1
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
86
1
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.
 

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