HELP!!Can you please explain too?Iwould really appreciate this!! 1) A 0.43-kg ma

Question

HELP!!Can you please explain too?Iwould really appreciate this!!

1) A 0.43-kg mass is hanging from a spring with spring constant 17 N/m. Then the mass is displaced from the equilibrium by 3.5 cm and let go. What is the resulting angular frequency of the oscillation?
(use rad/s as units)

2) A 0.65-kg mass is hanging from a spring with spring constant 11 N/m. Then the mass is displaced from the equilibrium by 2.7 cm and let go. What is the resulting frequency of the oscillation?

3) A 0.93-kg mass is hanging from a spring with spring constant 20 N/m. Then the mass is displaced from the equilibrium by 2.7 cm and let go. What is the resulting period of the oscillation?

4)An ideal massless spring is hanging vertically. When a 0.93-kg mass is attached to the spring, the spring stretches by 5 cm before reaching a new equilibrium. Then the mass is displaced by 1 cm from the new equlibrium. What is the frequency of the resulting oscillation?

Explanation / Answer

1) w = sqrt(k/m) = sqrt(17/0.43)=6.29 rad/s

2) f = 1/( 2 pi) sqrt(k/m) = 1/(2*pi) sqrt(11/0.65)=0.655 Hz

3) T = 2 pisqrt(m/k) = 2*pi*sqrt(0.93/20)=1.35 s

4) so we know F = kx

k = 0.93*9.81/5.0E-2=182.47

f = 1/(2 pi) sqrT(k/m) = 1/(2*pi) sqrt(182.47/0.93)=

2.23 Hz

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