A 3.90-kg object is attached to a spring and placed on frictionless, horizontal

Question

A 3.90-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 28.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations.

(a) Find the force constant of the spring.
N/m

(b) Find the frequency of the oscillations.
Hz

(c) Find the maximum speed of the object.
m/s

(d) Where does this maximum speed occur?
x =+- m

(e) Find the maximum acceleration of the object.
m/s2

(f) Where does the maximum acceleration occur?
x =+- m

(g) Find the total energy of the oscillating system.
J

(h) Find the speed of the object when its position is equal to one-third of the maximum value.
m/s

(i) Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value.

Explanation / Answer

a) K = F/delta X K = 28/0.2 = 140 N/m

b) omega = ( K/M)^0.5 = 6 so frequency = omega/2pi = 6/2*3.14 = 0.95Hz = 1Hz

c) maximum speed = amplitude * omega [ amplitude = max displacement = 0.2 m ]

= 0.2 * 6 = 1.2 m/sec

d) at origin [ X= 0 ]

e) max acceleration amplitude * omega ^2 = 0.2 * 6^2 = 7.2 m/sec^2

f) X = + 0.2 m 0R X = - 0.2 m

g) total energy = ( K * A^2)/2 = ( 140 * 0.2^2)/2 = 2.8 J

h) then sin( theta ) = 1/3 so theta = 19.47 degree

velocity = amplitude * omega * cos ( 19.47) = 1.13 m/sec

i) acceleration = amplitude * omega * sin( 19.47 ) = 2.399 = 2.4 m/sec^2

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