Question

A 1.80-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 15.0 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.
m/s2

Answer 1

Answer:

Explanation:

(a) Force constant = force/displacement = 15/0.2

=75 N/m

(b) For SHM with spring, the angular frequency = sqrt ( force constant / mass )

= sqrt ( 75 / 1.8 )

= 6.455

Frequency = Angular frequency / 2pi

= 1.03Hz

(c) Maximum speed is when object's potential energy is at minimum.

1/2*mass*speed^2 = 1/2*force constant*displacement^2

speed^2 = 75*0.2^2/1.8 = 1.67

speed = sqrt (1.67)

= 1.29 m/s

(d) It occurs at x=0

(e) Maximum acceleration = Force / m = 15/1.8

= 8.33 m/s^2

(f) It occurs at x=0.2m

Answer 2

Answer:

Explanation:

given:

The mass of the object attached to the spring: m = 1.80 kg

The magnitude of the maximum horizontal force exerted: F = 15.0 N

The amplitude of the spring-object oscillation: A = 0.200 m

a. k=15/.2=75N/m

b. (2pi*frequency)^2 = k/m = 75/1.8 = 41.67

frequency = 1.03Hz

c. max speed^2 = amplitude^2*k/m = 0.2^2 * 41.67

max speed = 1.29 m/s

d. x=0m

e. max accel = F/m = 15/1.8 = 8.33 m/s^2

f. x=0.2m

e. energy = 1/2*k*A^2

= 1/2*75*0.2^2

= 1.5J