If the net force acting on a particle is a linear restoring force, the motion wi
ID: 1983348 • Letter: I
Question
If the net force acting on a particle is a linear restoring force, the motion will be simple harmonic motion around the equilibrium position.The position as a function of time is x(t)=Acos(omega t + phi_0). The velocity as a function of time is v_x(t)=-omega A sin(omega t + phi_0). The maximum speed is v_{ m max} = omega A. The equations are given here in terms of x, but they can be written in terms of y, theta or some other parameter if the situation calls for it.
The amplitude A and the phase constant phi_0 are determined by the initial conditions through x_0=Acosphi_0 and v_{0x}=-omega A sinphi_0.
The angular frequency omega (and hence the period T = 2pi/omega) depends on the physical properties of the situtaion. But omega does not depend on A or phi_0.
Mechanical energy is conserved. Thus rac{_1}{^2}mv_x^2 + rac{_1}{^2}kx^2 = rac{_1}{^2}kA^2= rac{_1}{^2}m(v_{ m max})^2.Energy conservation provides a relationship between position and velocity that is independent of time.
A). The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where (t) is in seconds. Determine the velocity at t=0.40s.
B). Assume that the oscillating mass described in Part A is attached to a spring. What would the spring constant k of this spring be?
C). What is the total energy E of the mass described in the previous parts?
Explanation / Answer
a) Differentitate the position relation respect "t": x(t)=0.02Cos(10t) v(t)=-0.02Sen(10t)*10=-0.20Sen(10t) speed at t=0.40s is: v(0.40)=-0.20Sen(0.40*10) b) acceleration is: a(t)=-0.20Cos(10t)*10=-2Cos(10t) F=ma=-2mCos(10t)=-kx=-k0.02Cos(10t) -2mCos(10t)=-k0.02Cos(10t) -2m=-k0.02 k=-2m/0.02 c) E=0.5k0.02²