Hooke’s Law and the Simple Harmonic Oscillator

2springsA mass m is attached to a spring S and oscillates with a period T.
(a) What would be the period of the oscillations on a planet with surface gravity 3g?
(b) What would be the period of the oscillation if two springs S are connected in series or in parallel?

Solution.

The period of the oscillations are given by the formula

T = 2π*√(m/k)

hence, (a) the period of the oscillations on a planet with surface gravity 3g will be the same because period doesn’t depend on g.

(b)
Two springs in series.
The force is the same on each of the two springs. Therefore, using Hooke’s Law

F = -k1x1 = -k2x2                              (1)

Solving for x1 in terms of x2

x1 = k2/k1*x2                                      (2)

We are looking for the effective spring constant so that

F = -keff(x1+x2)                                  (3)

Equating (3) with the right side of (1) and substituting (2) gives

k2x2 = keff(k2/k1*x2+x2)

Divide through by x2 to obtain

k2 = keff(k2/k1+1)

or

keff = k1*k2/(k1+k2)

In the case of the two same springs,

keff = k/2

Hence, the period of the oscillation for the two springs S are connected in series is given by the formula

Tser = 2π*√(m/(k/2)) = √2T

Two springs in parallel.
The displacement of the spring is the same on each of the two springs. Therefore, using Hooke’s Law

F = -(k1x+k2x) = -keffx

Thus, the effecting spring constant is given by

keff = k1+k2

In the case of the two same springs,

keff = 2k

Hence, the period of the oscillation for the two springs S are connected in parallel is given by the formula

Tpar = 2π*√(m/2k) = T/√2

 

 

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