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what is the object’s velocity when its potential energy is 23e?

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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?

what is the object's velocity when its potential energy is 23e?
what is the object’s velocity when its potential energy is 23e?

Data

Revealed values from the question

Mass of the object = m

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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)kx2 = (1/2)kA2 [Because x = A from the question]

The kinetic energy of the spring is given by K.E = (1/2)mv2

E = P.E = (1/2)kA2

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)mv2

We will now make v subject of the formula

(1/2)mv2 = E – 2/3E

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

(1/2)mv2 = E (1 – 2/3)

to get

(1/2)mv2 = (1/3)E

But E = (1/2)kA2 and this implies that the above equation is

(1/2)mv2 = (1/3)(1/2)kA2

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

mv2 = (1/3)kA2

We can now divide both sides by m to get

v2 = (1/3m)kA2

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

v = √(kA2/3m)

Hence the answer to what is the object’s velocity when its potential energy is 23e? is v = √(kA2/3m)

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Sources:

Wikipedia

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