here,
8)
let the spring constant be K
when m1 = 5 g = 0.005 kg hangs from the spring
x = 3 cm = 0.03 m
equating the forces vertically
K * x = m * g
K * 0.03 = 0.005 * 9.81
solving for K
K = 1.635 N/m
New mass attached , m2 = 50 g = 0.05 kg
the period of motion , T = 2*pi*sqrt(m/K)
T = 2*pi*sqrt(0.05 /1.635)
T = 1.1 s
8. A spring stretches 3 cm when a 5 g object is hung from it. The...
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