Find the maximum and minimum velocity attained by the oscillating glider of Exam

Question

Find the maximum and minimum velocity attained by the oscillating glider of Example 14.2 in your textbook (page 442). (b) Find the maximum and minimum accelerations. (c) Find the velocity vx and acceleration ax when the glider is halfway from its initial position to the equilibrium position x = 0. (d) Find the potential energy, kinetic energy and total energy at this halfway position. Initial displacement x= +0.015 m , initial velocity v= +0.40 m/s.

(ex. 14.2)

A spring is mounted horizontally, with its left end fixed. A spring balance attached to the free end and pulled toward the right indicates that the stretching force is proportional to the displacement, and a force of 6.0 N causes a displacement of 0.030 m. we replace the spring balance with a 0.50 kg glider, pull it 0.020 m to the right along a frictionless air track, and release it from rest. (a) Find the force constant k of the spring. (b) find the angular frequency w, frequency f, and period T of the resulting oscillation.

example 14.2 aswers:

m= 0.50 kg

k= -200 N/m

w= 20 rad/s

f= 3.2 Hz

T= 0.31s

Explanation / Answer

Part a

Maximum velocity of oscillating glider Vmax=wx0

Where x0 is maximum displacement from mean position

Vmax =20*0.02=0.40 m/s

Minimum velocity of oscillating glider Vmin=-wx0

Vmax =-20*0.02= -0.40 m/s

Part b

Maximum acceleration of oscillating glider Amax=w2x0

                                                                                = 20*20*0.02=8 m/s^2

Minimum acceleration of oscillating glider Amin= -w2x0

                                                                                                                        = -20*20*0.02 =-8 m/s^2

Part c

Velocity Vx=-w(x02 – x2 )1/2

At x=x0 /2

                Vx =-wx0 (¾)1/2 = -0.3464 m/s

Acceleration Ax =-w2x

                Ax = -20*20*0.01 = -4 m/s^2

Part d

Kinetic energy

                K.E.=(mv^2 )/2

At x= x0/2=0.01 m ,    v=-0.3464 m/s

                K.E.=0.029998 J

Potential energy

                P.E.= kx2 /2

At x= x0/2=0.01 m ,

                P.E.= -0.01 J

Total energy= potential energy + kinetic energy

                         = 0.029998 + (-0.01)

                                =0.019998 J

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