Question
A 380 g block connected to a light spring for which the force constant is 8.90 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.30 cm from equilibrium and released from rest as in the figure.
(a) Express the position, velocity, and acceleration as functions of time.
Find the phase constant from the initial condition that x = A at t = 0:
x(0) = A cos = A --> = 0
Use the equation to write an expression for x(t):
x = Acos(t + ) = ?
Use the equation to write an expression for v(t):
v = -Asin(t + ) = ?
Use the equation to write an expression for a(t):
a = -^2 Acos(t + ) = ?
Explanation / Answer
Given: Mass of the block = M = 380 g = 380 x10-3 kg Force constant = K = 8.9 N /m Thus,Angular frequency of the osscillation ; w = K / M = [8.9 / 380 x10-3 ] = 4.839 rad/s Since , the block is displaced 5.30 cm from equilibrium and released from rest Thus, Amplitude of the oscillation : A = 5.3 cm = 0.053 m Let , Displace ment of the oscillation: x = Acos(t + ?) = (0.053) cos (4.839 t + ) Since it is realesed feom rest , x= A at t= 0 thus, intial phase = 0 so, displacement = x = (0.053) cos (4.839 t ) velocity = V = dx /dt = 0.053(4.839) [- sin(4.839 )t] = - (0.2564 )sin(4.839 )t acceleration = a = dV /dt =- (0.2564 )(4.839)cos(4.839 )t = - ( 1.241) cos(4.839 )t Mass of the block = M = 380 g = 380 x10-3 kg Force constant = K = 8.9 N /m Thus,Angular frequency of the osscillation ; w = K / M = [8.9 / 380 x10-3 ] = 4.839 rad/s Since , the block is displaced 5.30 cm from equilibrium and released from rest Thus, Amplitude of the oscillation : A = 5.3 cm = 0.053 m Let , Displace ment of the oscillation: x = Acos(t + ?) = (0.053) cos (4.839 t + ) Since it is realesed feom rest , x= A at t= 0 thus, intial phase = 0 so, displacement = x = (0.053) cos (4.839 t ) velocity = V = dx /dt = 0.053(4.839) [- sin(4.839 )t] = - (0.2564 )sin(4.839 )t acceleration = a = dV /dt =- (0.2564 )(4.839)cos(4.839 )t = - ( 1.241) cos(4.839 )t