Question

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is

Answer 1

Answer:

The kinetic energy at a displacement of half the amplitude is 37.5 J

Explanation:

Given;

total energy on the spring, E = 50 J

When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.

E = K + U

Where;

K is the kinetic energy

U is the elastic potential energy

K = E - U

K = E - ¹/₂KA²

When the displacement is half = ¹/₂(A) = A/₂

K = E - ¹/₂K(A/₂)²

K = E - ¹/₂K(A²/₄)

K = E - ¹₄(¹/₂KA²)

Recall, E = ¹/₂KA²

K = ¹/₂KA² - ¹₄(¹/₂KA²)     (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)

K = 1(¹/₂KA²) - ¹₄(¹/₂KA²)  = ³/₄(¹/₂KA²)

K = ³/₄(¹/₂KA²)

But E = ¹/₂KA² = 50J

K = ³/₄ (50J)

K = 37.5 J

Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J

Answer 2

The kinetic energy when the displacement is half the amplitude

Given the following data:

• Total energy = 50 Joules.

• Displacement, x = $\frac{A}{2}$

To find the kinetic energy when the displacement is half the amplitude:

The total energy of the system of a block and a spring is the sum of the spring's elastic potential energy and kinetic energy of the block and it's proportional to the square of the amplitude.

Mathematically, the total energy of the system of a block and a spring is given by the formula:

$T.E = U + K.E$   .....equation 1.

$T.E = \frac{1}{2} kA^2$

Where:

• T.E is the total energy.

• U is the elastic potential energy.

• K.E is the kinetic energy.

• A is the amplitude.

Making K.E the subject of formula, we have:

$K.E = T.E - U$   .....equation 2.

But, $U = \frac{1}{2} kx^2$    ....equation 3.

Where:

• k is spring constant.

• x is change in position (displacement).

Substituting the eqn 3 into eqn 2, we have:

$K.E = T.E - \frac{1}{2} kx^2$

$K.E = T.E - \frac{1}{2} k(\frac{A}{2})^2\\\\K.E = T.E - \frac{1}{2} k(\frac{A^2}{4})\\\\K.E = T.E - \frac{1}{4} (\frac{1}{2} kA^2)\\\\K.E = T.E - \frac{1}{4} (T.E)\\\\K.E = 50 - \frac{1}{4} (50)\\\\K.E = 50 - 12.5$

K.E = 37.5 Joules.

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