Question

a body mass m = 2.025 kg is suspended by an ideal spring. At equilibrium the spring elongation is y0 = 2.5 cm. The upper end of the spring is subjected to a harmonic oscillation movement with amplitude A = 1 mm. The oscillator quality factor is Q = 15. Determine:
a) the own pulsation of the oscillator
b) the amplitude of the forced oscillations when the oscillator's own pulse is equal to the pulsation of the external force

Answer 1

The quality factor of the oscialltor Q=15

Spring harmonic oscillation amplitude A=1 mm = 0.1 cm = 0.001 m

Spring elongation y0=2.5 cm = 0.025 m

mass of the body m = 2.025 kg

spring constant k = Force/elongation

k = mass X acceleration due to gravity / elongation

a) own pulsation of the oscillator is nothing but the frequency of the oscillator.

frequency

b) Finding the amplitude of the forced oscillations when the oscillator's own pulse is equal to the pulsation of the external force

time period of the oscillator T = 1/f

T = 0.3172 sec

from harmonic oscillator equation, Amplitude

where A is amplitude

X is displacement

w is the oscillator frequency i.e., w=2pi f

phi is phase difference.

here, in this case the phase difference is zero as the external force is equal to the pulsation of the oscillator.

Therefore, the amplitude