A mass m = 2.67 kg is attached to a spring of force constant k = 50.9 N/m and se

Question

A mass m = 2.67 kg is attached to a spring of force constant k = 50.9 N/m and set into oscillation on a horizontal frictionless surface by stretching It en amount A = 0.15 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle has a constant counterclockwise angular speed omega, constant tangential speed V = omega A, and centripetal acceleration of constant magnitude ac = omega 2A. Determine the following. maximum speed of the oscillating mass m/s magnitude of the maximum acceleration of the oscillating mass m/s2 magnitude of the maximum force experienced by the oscillating mass N maximum kinetic energy of the oscillating mass J maximum elastic potential energy of the spring attached to the mass J total energy of the oscillating mass-spring system J If the record of time starts when x = +A and v = 0, determine expressions for the displacement, velocity, and acceleration of the oscillating mass along the x-axis at any time t later (Your expression should be in terms of the variable

Explanation / Answer

omega = (k/m)^0.5 = sqrt(19.063) = 4.36

initial potential energy stored in the system when the spring is sreched by 0.15 m

= 1/2*k*0.15^2 = 0.57 J

by consrvation of energy

this potential energy will be converted nto kinetic energy at equillibrium position where it will have maximum kintic enrgy

and hence maximum velocity

so

0.57 = 1/2*2.67*v^2

v = 0.65 m/s

maximum speed = 0.65 m/s

acceleration = w^2 *displacement

maximum acceleration = w^2 *A as A is the maximum displacement

maximum acceleration = 19.063* 0.15 = 2.85 m/s^2

maxmum force will be exerted when maximum acceleration is taking place

maximum force = 2.67*2.85 = 7.63 N

maximum kinretic energy has already been found as 0.57 J

similarly maximum potential energy = 0.57 J

total energy of oscillation is same as maximum kinetic energy by conservation of enrgy principle

so its also 0.57J

at t= 0

displacement = A

the functon defining equation of positon be A cos(w*t+phi) since it is simple harmonic motion it can be eigther sin or cos function any function will be correct

so we have chosen cos function

A is the maximum displacement because it is initial displacement

x(t) = Acos(wt+phi)

at t= 0

A = Acos(phi)

cos(phi) = 1

phi = 0

At t = 0

v(t ) = 0

v(t) =w A sin(wt)

0 = w*A sin(0)

lhs = rhs so our equation is correct

x(t) = A *cos(4.36*t)

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