A mass m = 1.37 kg is attached to a spring of force constant k = 46.9 N/m and se

Question

A mass m = 1.37 kg is attached to a spring of force constant k = 46.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.17 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 ?, constant tangential speed V = ?A, and centripetal acceleration of constant magnitude ac = ?2A.

(a) Determine the following.

A mass m = 1.37 kg is attached to a spring of force constant k = 46.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.17 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 ?, constant tangential speed V = ?A, and centripetal acceleration of constant magnitude ac = ?2A. Determine the following. maximum speed of the oscillating mass magnitude of the maximum acceleration of the oscillating mass magnitude of the maximum force experienced by the oscillating mass maximum kinetic energy of the oscillating mass maximum elastic potential energy of the spring attached to the mass total energy of the oscillating mass-spring system 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 t and other numerical values. Assume any numerical values in your expression are in standard SI units, but do not enter units into your expression.) x = ? v = ? a = ? If the record of time starts when x = 0 and v = +?A, 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 t and other numerical values. Assume any numerical values in your expression are in standard SI units, but do not enter units into your expression.) x = ? v = ? a = ?

Explanation / Answer

a)

1. mass will have max. speed when it will be passing through equilibrium position,

so,

by conservation of energy

1/2*m*Vmax^2 = 1/2*k*A^2

Vmax = sqrt(k*A^2/m) = sqrt(46.9*0.17^2/1.37) = 0.995 m/s

2.

Fmax= mamax = k*A

amax = k*A/m = 46.9*0.17/1.37 = 5.82 m/s^2

3.

Fmax = k*A = 46.9*0.17 = 7.973 N

4.

Kmax = 1/2*m*Vmax^2 = 1/2*1.37*0.995^2 = 0.678 J

5.

Pmax = 1/2*k*A^2 = 1/2*46.9*0.17^2 = 0.678 J

6.

Total energy = 1/2*k*A^2 = 1/2*46.9*0.17^2 = 0.678 J

b)

w = sqrt(k/m) = sqrt(46.9/1.37) = 5.85 rad/s

For Simple harmonic Motion , we have

x=A*cos(wt+P) .....(P=phase)

at t=0 ==> x=A

so,

1 = cos(P) ...

P = 0 degree = 0 rad

so,

our equation is

x = A*cos(wt+P) = 0.17*cos(5.85*t)

V = dx/dt = -A*w*sin(w*t) = -0.17*5.85*sin(5.85*t) = -0.9945*sin(5.85*t)

a = dV/dt = -0.9945*5.85*cos(5.85*t) = -5.82*cos(5.85*t)

so,

x = 0.17*cos(5.85*t)

V = -0.9945*sin(5.85*t)

a = -5.82*cos(5.85*t)

c)

For Simple harmonic Motion , we have

x=A*cos(wt+P) .....(P=phase)

at t=0 ==> x=0

so,

0 = cos(P) ...

P = 90 degree = pi/2 rad

so,

our equation is

x = A*cos(wt+P) = 0.17*cos(5.85*t+pi/2)

V = dx/dt = -A*w*sin(w*t) = -0.17*5.85*sin(5.85*t) = -0.9945*sin(5.85*t+pi/2)

a = dV/dt = -0.9945*5.85*cos(5.85*t) = -5.82*cos(5.85*t+pi/2)

so,

x = 0.17*cos(5.85*t+pi/2)

V = -0.9945*sin(5.85*t+pi/2)

a = -5.82*cos(5.85*t+pi/2)

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