# Block Oscillating on Spring

## 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?

supratim1
Gold Member
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.

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.

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.