## Question

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. What is the object’s velocity when its potential energy is 2/3E?

### Data

**Revealed values from the question**

Mass of the object = m

Spring Constant = k

Maximum displacement of the spring, x = A

Mechanical Energy of the system, M.E = E

Potential energy, P.E = 2/3E

**Unrevealed Value** **from the question:**

The velocity of the object = v

**Important Formulae to use**:

**Mechanical energy is given** by M.E = Potential Energy (P.E) + Kinetic Energy (K.E)

**And the potential energy of an elastic spring is** P.E = (1/2)kx^{2} = (1/2)kA^{2} [Because x = A from the question]

**The kinetic energy of the spring** **is given by** K.E = (1/2)mv^{2}

E = P.E = (1/2)kA^{2}

### Solution

We will now apply our data and formulae to solve the problem

Since M.A = P.E + K.E

We can equally insert our data into the above formula for mechanical energy of the system

E = 2/3E + (1/2)mv^{2}

We will now make v subject of the formula

(1/2)mv^{2} = E – 2/3E

solving the right-hand side we will substract 2/3E from E and obtain

(1/2)mv^{2} = E (1 – 2/3)

to get

(1/2)mv^{2} = (1/3)E

But E = (1/2)kA^{2} and this implies that the above equation is

(1/2)mv^{2} = (1/3)(1/2)kA^{2}

from the above equation, 1/2 will cancel each other on both hands leaving us with

mv^{2} = (1/3)kA^{2}

We can now divide both sides by m to get

v^{2} = (1/3m)kA^{2}

We can now take the square-root of both sides to make v subject of the formula

v = √(kA^{2}/3m)

**Hence the answer to what is the object’s velocity when its potential energy is 23e? is v = √(kA ^{2}/3m)**

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