A mass attached to the end of a spring forms a simple oscillator. The mass and s

Question

A mass attached to the end of a spring forms a simple oscillator. The mass and spring oscillate in a horizontal plane so that we can ignore gravity. When mass M is attached to a spring of force constant 250 N/m it oscillates at a frequency of 2.5 Hz. The amplitude of oscillation is 5.0 cm. What is the value of M to cause the above oscillation? What is the force on M when it is displaced from its equilibrium position by 1/2 of its maximum displacement in its oscillation? An elastic cord is 80. cm long when it is supporting a mass of 10 kg hanging from it at rest at rest. When an additional 4.0 kg is added, the cord is 82.5 cm long. What it the spring constant of the cord? (1600 N/m) What is the length of cord when no mass is hanging from it? (74 cm)

Explanation / Answer

a)we know that for a horizontal mass spring system,

w = sqrt(k/m)

2 pi f = sqrt (k/m)

4 pi^2 f^2 = k/m

m = k/4 pi^2 f^2

m = 250/4 x 3.14^2 x 2.5^2 = 1.014 kg

Hence, m = 1.014 kg

x = 1/2A = 1/25 = 2.5 cm = 0.025 m

F = -kx = -250 x 0.025 = -6.25 N

Hence, F = -6.25 N

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