A 27.0 kg block at rest on a horizontal frictionless air track is connected to t

Question

A 27.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 5.20 s. What is the position of the mass 4.264 s after the mass is released? (in m)

A 27.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 5.20 s. What is the position of the mass 4.264 s after the mass is released? (in m)

Explanation / Answer

= 2/T = 1.2083

general solution of a harmonic oscillator;

x = Acos t + Bsint

given at t=0;

x= 0.350

so, A = 0.350m

at t = 0s

also dx/dt = 0

Bwcos 0 = 0

B = 0

So, x = 0.35cos(1.2083t)

So, at t = 4.264s

x = 0.35cos(1.2083*4.264) = 0.149m

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