A4.00-kg object is attached to a spring and placed on frictionless, horizontal surface. a horizontal force of 27.0 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/s^2

a)

Nm⁻¹

b)

Hz

c)

ms⁻¹

d)

m

e)

ms⁻²

f)

m

Explanation:

a)

= force required to hold the object at rest connected with stretched spring = 27 N

= stretch in the spring from equilibrium position = 0.2 m

= force constant of the spring

force required to hold the object at rest is same as the spring force , hence

inserting the values

Nm⁻¹

b)

frequency of the oscillations is given as

inserting the values

Hz

c)

= Amplitude of oscillations = 0.2 m

= angular frequency

Angular frequency is given as

rads⁻¹

Maximum speed of oscillation is given as

ms⁻¹

d)

maximum speed of the object occurs at the equilibrium position, hence

m

e)

Maximum acceleration of oscillation is given as

ms⁻²

f)

maximum acceleration occurs when the object is at extreme positions, hence

m

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.