An undamped 1.23 kg horizontal spring oscillator has a spring constant of 37.4 n/m. while oscillating, it is found to have a speed of 2.48 m/s as it passes through its equilibrium position. what is its amplitude of oscillation?
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Ответ:
0.45 m
Explanation:
m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s
velocity is maximum when it passes through the mean position.
T = 1.139 sec
w = 2 π / T
w = 2 x 3.14 / 1.139
w = 5.51 rad / s
Vmax = w A
Where, A be the amplitude
2.48 = 5.51 A
A = 2.48 / 5.51 = 0.45 m
Ответ:
Choose the 2nd Option