physics-mechanics

a. If the two springs are connected in parallel, then the equivalent spring constant is given by:

k = k1 + k2

The system is undergoing simple harmonic motion (the proof is skipped), the period of the oscillation is given by,

T = 2π√(m/k)

= 2π√[m / (k1 + k2)]

So, frequency, f

= 1 / T

= (1/2π)√[(k1 + k2) / m]



b. If the two springs are connected in series, the equivalent spring constant is given by:

1/k = 1/k1 + 1/k2

So, k = k1k2 / (k1 + k2)

The system is undergoing simple harmonic motion (the proof is skipped), the period of the oscillation is given by,

T = 2π√(m/k)

= 2π√[m(k1 + k2) / k1k2]

So, frequency, f

= 1 / T

= (1/2π)√[k1k2 / m(k1 + k2)]

2009-03-05 07:39:19 補充：
I have checked. I haven't mixed them up. I am sure my answer is correct. It is the opposite of that of resistors in the circuit. That is spring in parallel is equivalent to resistor in series, and vice versa.

thx for answering but i think u mixed up in series and in parallel

2009-03-05 20:14:41 補充：
thanks i think you are right now