# If it is pulled down from the at rest position by 40

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If it is pulled down from the “at rest” position by 4.0 cm and released, determine: (i) The maximum speed attained by the mass, (ii) The maximum acceleration attained by the mass, and (iii) The number of oscillations it makes in 60 sec ANS: We recognize this as a harmonic oscillator for which x ( t ) = A cos( ω t + φ ) ω = k m with The second of these will give us the frequency and hence the answer to (iii). For (i) and (ii) we need the amplitude of the velocity and acceleration oscillations: v ( t ) = dx ( t ) dt = - A ω sin( ω t + φ ) and a ( t ) = dv ( t ) dt = - A ω 2 cos( ω t + φ ) So v max = A ω a max = A ω 2 and To complete the problem we must determine A , the amplitude of teh x(t) oscillation.

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Phys 2A - Mechanics Example: A 5.0 kg mass hangs at rest from a 20 N/m spring. If it is pulled down from the “at rest” position by 4.0 cm and released, determine: (i) The maximum speed attained by the mass, (ii) The maximum acceleration attained by the mass, and (iii) The number of oscillations it makes in 60 sec We have initial conditions x(0) = x 0 = 4.0 cm, and v(0)=0 (it is “released”): x (0) = A cos( φ ) and v (0) = dx ( t ) dt 0 = - A ω sin( φ ) which are solved by A = x 0 (and, although we won’t need it, ). φ = 0 ANS (cont’d) Since ω appears in the solution to each part, let’s compute it: (i) and (ii) ω = k m = 20 5 . 0 rad/s = 2 . 0 rad/s v max = x 0 ω = (0 . 04)(2 . 0) m/s = 0 . 08 m/s a max = x 0 ω 2 = (0 . 04)(2 . 0) 2 m/s = 0 . 16 m/s 2
Phys 2A - Mechanics And, last, (iii) (which we could have done first) The frequency, giving the number of oscillations per second is f = ω 2 π = 2 . 0 2 π s - 1 = 0 . 32 s - 1 The number of oscillations in a time interval is Δ t N of oscillations = f Δ t = (0 . 32)(60) = 19

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Phys 2A - Mechanics Example: The physical pendulum ANS: In lecture 24 we found the equation of motion axis CG W θ d 2 θ dt 2 = - Mg I sin θ Compare with SHO d 2 x dt 2 + ω 2 x = 0 Recall that in general, this applies only to small oscillations. This means θ is small, so we can approximate (Taylor expansion): sin θ θ (sin θ = θ - 1 3!
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