Question

An object of mass 2 kg hangs from a spring of negligible mass. The sping is extended by 2.5 cm when the object is attached. The top end of spring is oscillated up and down in simple harmonic motion with and amplitude of 1 mm. The quality factor, Q, of the system is 15.

a. What is Wo (undamped frequency) for the system?

b. What is the amplitude of forced oscillation at W = Wo

c. What is the mean power input to maintain the forced oscillation at a frequency 2% greater than Wo?

Answer 1

(a) We know that

F = kx where F = mg

therefore,

k = mg/x

let's assume g to be 10 m/s²

k = 2*10 / 2.5e-2

k = 800 N/m

Now,

_o = sqrt (k/m)

_o = sqrt (800 / 2)

_o = 20 rad/sec

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(b) amplitude of forced oscillation at = _o is

A = F_o / m_o D

where D is damping of system

F_o = 800 * 1e-3 = 0.8 N

D = _o / Q where Q is quality factor

D = 20 / 15 = 1.333

Therefore,

A = 0.8 / 1.333*20*2

A = 1.5 cm

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(C) Here

= _o + 0.02_o

= 20 + 0.02*20

= 20.4 rad/sec

Mean power output is given as

P = 0.5*F_o²²D / m * (_o² - ²)² + ²D²

P = 0.0887 W