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A 3.94-kg block extends a spring 15.7 cm from its unstretched position. The block is removed and a 0.520-kg object is hung from the same spring. Find the period of its oscillation.

A 3.94-kg block extends a spring 15.7 cm from its unstretched position. The block is removed and a 0.520-kg object is hung from the same spring. Find the period of its oscillation.

Grade:11

2 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
To find out the force constant k, substitute 3.94 kg for mass of the block m, 9.81 m/s2 for g and 15.7 cm for x in the equation k = mg/x,
k = mg/x
= (3.94 kg) (9.81 m/s2) / (15.7 cm)
= (3.94 kg) (9.81 m/s2) / (15.7 cm×10-2 m/1 cm)
= (0.246 kg. m/s2) / m
= (0.246 kg. m/s2× (1 N/1 kg.m/s2)) / m
= 0.246 N/m
To obtain the period T of oscillation, substitute 0.246 N/m for k and 0.520 kg for mass of the object M in the equation T = 2π √M/k,
T = 2π √M/k
= 2×3.14×√(0.520 kg)/( 0.246 N/m)
= 0.289 √kg/(N/m) (1 kg.m/s2 /1 N)
= 0.289 s
From the above observation we conclude that, the period of oscillation would be 0.289 s.
pa1
357 Points
6 years ago
/1 N)= 0.289 sFrom the above observation we conclude that, the period of oscillation would be 0.289 s.2)) / m = 0.246 N/mTo obtain the period T of oscillation, substitute 0.246 N/m for k and 0.520 kg for mass of the object M in the equation T = 2π √M/k,T = 2π √M/k = 2×3.14×√(0.520 kg)/( 0.246 N/m) = 0.289 √kg/(N/m) (1 kg.m/s2× (1 N/1 kg.m/s2) / m = (0.246 kg. m/s2 m/1 cm) = (0.246 kg. m/s-2) / (15.7 cm×102) / (15.7 cm) = (3.94 kg) (9.81 m/s2 for g and 15.7 cm for x in the equation k = mg/x,k = mg/x = (3.94 kg) (9.81 m/s2To find out the force constant k, substitute 3.94 kg for mass of the block m, 9.81 m/s

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