6bpracsol[1]

# 6bpracsol[1] - Solutions to Practice Midterm Problem 1 For...

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Unformatted text preview: Solutions to Practice Midterm Problem 1: For part 1 , the differential equation governing the motion of the system is- kx = m d 2 x dt 2 (1) which can also be written d 2 x dt 2 + ω 2 x = 0 , where ω = r k m (2) Part 2: The general solution to this equation is x ( t ) = A cos( ωt + φ ) v ( t ) = dx dt =- ωA sin( ωt + φ ) (3) Applying the initial conditions, we have x (0) =- A = A cos φ v (0) = 0 =- ωA sin φ (4) From the first equation, we see that cos φ =- 1 φ = π (5) The solution satisfying the initial conditions is therefore x ( t ) = A cos( ωt + π ) =- A cos( ωt ) (6) Part 3: We verify that our solution satisfies the differential equation as follows: v ( t ) = dx dt = ωA sin( ωt ) a ( t ) = d 2 x dt 2 = ω 2 A cos( ωt ) (7) We now have d 2 x dt 2 + ω 2 x = ω 2 A cos( ωt )- ω 2 A cos( ωt ) = 0 (8) That is, we have verified that 1 d 2 x dt 2 + ω 2 x = 0 d 2 x dt 2 + k m x = 0 m d 2 x dt 2 + kx = 0- kx = m d 2 x dt 2 (9) Part 4: The solution x ( t ) represents a mass m oscillating with angular frequency ω between positions x =- A and x = A , with x = 0 as the equilibrium position. A time T = 2 π/ω is taken for the mass to complete each cycle, i.e., to return to its initial position. The initial position is at x =- A , which is the mass’s most negative position. The spring provides the restoring force- kx that interacts with the inertia of the mass to produce the oscillations. Part 5: The velocity and acceleration are given by (7), above: v ( t ) = dx dt = ωA sin( ωt ) a ( t ) = d 2 x dt 2 = ω 2 A cos( ωt ) (10) The speed is the magnitude of the velocity: | v ( t ) | = | ωA sin( ωt ) | (11) The angular frequency ω , as stated above, is ω = r k m (12) Parts 7-8): As determined above, φ = π . The solution x ( t ) is x ( t ) = A cos ˆ r k m t + π ! =- A cos ˆ r k m t ! (13) Part 9: The maximum speed is v max = ωA , because this is the amplitude of v ( t ). This occurs at x = 0, i.e., when the mass is at equilibrium. This can be verified using energy conservation: 2 1 2 kx 2 + 1 2 mv 2 = 1 2 kA 2 x 2 + v 2 ω 2 = A 2 x ( v ) = r A 2- v 2 ω 2 (14) So, we have x ( v max ) = r A 2- v 2 max ω 2 = r A 2- A 2 ω 2 ω 2 = 0 (15) Part 10: The maximum acceleration is the maximum value of a ( t ) = ω 2 A cos( ωt ) This is given by a max = ω 2 A . To find the position where this occurs, we use the relationship a =- ω 2 x x ( a ) =- a ω 2 We therefore have x ( a max ) =- a max ω 2 =- ω 2 A ω 2 =- A Maximum acceleration occurs at x =- A . However, the acceleration has its maximum magnitude at x = A or x =- A . Part 11: The potential energy at time t is U ( t ) = 1 2 kx ( t ) 2 = 1 2 kA 2 cos 2 ( ωt ) (16) The kinetic energy at time t is K ( t ) = 1 2 mv ( t ) 2 = 1 2 mω 2 A 2 sin 2 ( ωt ) = 1 2 kA 2 sin 2 ( ωt ) (17) 3 The total energy at time t is therefore E ( t ) = U ( t ) + K ( t ) = 1 2 kA 2 cos 2 ( ωt ) + 1 2 kA 2 sin 2 (...
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6bpracsol[1] - Solutions to Practice Midterm Problem 1 For...

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