Question

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 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)

Answer 1

Incomplete question as we have not told to find any quantity so I have chosen  some quantities to find.So complete question is here

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 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.

Find (a) the force constant of the spring

 (b) the frequency of the oscillations

 (c) the maximum speed of the object.

Answer:

(a) $k=100N/m$

(b) $f=1.126Hz$

(c) $v_{max}=1.41m/s$

Explanation:

Given data

Mass of object m=2.00 kg

Horizontal Force F=20.0 N

Distance of the object from equilibrium A=0.2 m

To find

(a) Force constant of the spring k

(b) Frequency F of the oscillations

(c) The Maximum Speed V of the object

Solution

For (a) force constant of the spring k

From Hooke's Law we know that:

$F=kx\\k=F/x\\where\\x=A(amplitude)\\So\\k=F/A\\k=(20N/0.2000m)\\k=100N/m$

For (b) frequency F of the oscillations

The frequency of the motion for an object in simple harmonic motion is expressed as:

$f=\frac{1}{2\pi } \sqrt{\frac{k}{m} }\\ f=\frac{1}{2\pi } \sqrt{\frac{100N/m}{2.0kg} }\\f=1.126Hz$

For (c) maximum speed V of the object

For an object in simple harmonic motion the maximum values of the magnitude velocity is given as:

$v_{max}=wA\\ where\\w=\sqrt{\frac{k}{m} }\\ so\\v_{max}=\sqrt{\frac{k}{m} }A\\v_{max}=\sqrt{\frac{100N/m}{2.0kg} }(0.200m)\\v_{max}=1.41m/s$