Understanding Simple Harmonic Motion A mass m = 0.25 kg attached to a spring osc
ID: 2298334 • Letter: U
Question
Understanding Simple Harmonic Motion A mass m = 0.25 kg attached to a spring oscillates in the x direction on a frictionless horizontal surface. The amplitude of the oscillation is A = 1.50 cm, and at t = 0 the mass is at the position x = 1.00 cm. The position as a function of time, x(t), between -1.00 s and +1.00 s is plotted as a function of time in the following graph: Estimate the period, angular frequency to and the spring constant, k, from this plot. What is the total energy of the oscillation? Complete the specification of the equation for the position as a function of time, by finding the phase angle that is necessary to match the plot above. What are the velocity and the acceleration of the mass at t = 0.00 s?Explanation / Answer
From the graph, amplitude = height of peak = 1.5 cm
Time period T = 0.5 seconds approximately
Angular frequency w= 2pi/T = 2pi/0.5 = 4pi rad/s
angular frequency w = sqrt(k/m) => k = m*w^2 = 0.25 * 4pi*4pi = 39.5 N/m
Energy of oscillation = energy of the system = maximum potential energy stored in spring = 0.5*k*x^2 where x is amplitude of oscillation
Energy of oscillation = 0.5*39.5*(0.015)*(0.015)=0.0044 N.m
x(t) = 0.015 cos(4pi*t + phi) where phi is angular phase
x(0) = 0.01
=> cos(4pi*0+phi)=0.01/0.015=2/3
phi=arccos(2/3) = 0.84 rad
THerefore, x(t) = 0.015 cos(4pi*t + 0.84)
v(t) = x'(t) = 0.06pi*sin(4pi*t + 0.84)
v(0)=0.06pi*sin(4pi*0 + 0.84)=0.06pi*sin(0.84)=0.14 m/s
a(t)=v'(t)=-0.24*pi*pi*cos(4pi*t + 0.84)
a(0)=-0.24*pi*pi*cos(4pi*0 + 0.84)=-0.24*pi*pi*cos(0.84)=-1.58 m/s^2
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