Question

A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 18.0 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

Part A.

We know that spring force is given by:

|F| = k*x

given that when object is 0.200 m from equilibrium 18.0 N force is applied, So

k = |F|/x = 18.0/0.200 = 90 N/m

Part B.

We know that angular frequency in SHM motion is given by:

w = sqrt (k/m)

w = sqrt (90/2.20) = 6.396 rad/sec

Also, w = 2*pi*f, So frequency will be:

f = (1/(2*pi))*sqrt (k/m)

m = mass of object = 2.20 kg

So,

f = (1/(2*pi))*sqrt (90/2.20)

f = 1.01796 Hz = 1.02 Hz

Part C.

Max speed of object in SHM is given by:

V_max = A*w

A = Amplitude of motion = 0.200 m

So,

V_max = 0.200*6.396

V_max = 1.28 m/s

Part D.

Maximum speed of object in SHM occurs at the equilibrium points, which is at x = 0 m

x = 0 m

Part E.

Max acceleration of object in SHM is given by:

a_max = A*w^2

Using given values:

a_max = 0.200*(6.396)^2

a_max = 8.18 m/s^2

Part F.

max acceleration of object is occurs at the end points of motion, So at

x = 0.200 m

Part G.

total energy of oscillating system is given by:

TE = (1/2)*k*A^2

TE = (1/2)*90*0.200^2

TE = 1.8 J

Part H.

Speed of object in SHM at 'x' distance from equilibrium point is given by:

V = w*sqrt (A^2 - x^2)

Now we need speed when x = A/3, So

V = 6.396*sqrt (0.200^2 - (0.200/3)^2)

V = 1.21 m/s

Part I.

Acceleration of object at 'x' distance from equilibrium in SHM is given by:

a = w^2*x

Here, x = A/3, So

a = 6.396^2*(0.200/3)

a = 2.73 m/s^2