Asimple harmonic oscillator consists of a block of mass 45 g attached to a spring of spring constant 240 n/m, oscillating on a frictionless surface. if the block is given an initial velocity of 3.5 m/s starting at the equilibrium position, what is the maximum displacement of the block from its equilibrium position

The maximum displacement ,A= 0.047 m

Explanation:

Given that

Mass, m = 45 g= 0.045 kg

spring constant ,K= 240 N/m

Initial velocity at equilibrium ,v= 3.5 m/s

Lets take the maximum displacement is A

As we know that when the mass reached at the extreme position then the velocity of the mass will become zero.

From energy conservation

Now by putting the values

A=0.047 m

The maximum displacement ,A= 0.047 m

x = 4.79 cm

Explanation:

given,

mass of block = 45 g

spring constant = 240 N/m

initial velocity = 3.5 m/s

maximum displacement = ?

using conservation of energy

loss of KE = grain in spring constant

x = √0.00229

x = 0.0479 m

x = 4.79 cm

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation: