Problem 57 - University Physics with Modern Physics

Question

A mass \(m\) is attached to a spring with a spring constant of \(k\) and set into simple harmonic motion. When the mass has half of its maximum kinetic energy, how far away from its equilibrium position is it, expressed as a fraction of its maximum displacement?

Step-by-Step Solution

Answer

Answer: The distance from the equilibrium position when the mass has half of its maximum kinetic energy is approximately 0.707 times the maximum displacement.

Step 1: Write the equation for total energy in simple harmonic motion

For an object in simple harmonic motion, the total energy (E) is the sum of its kinetic energy (K) and potential energy (U). The total energy can be expressed as: E = K + U

Step 2: Write down the expressions for kinetic and potential energy

Kinetic energy K can be expressed as: K = \frac{1}{2}mv^2 where v is the velocity of the mass. Potential energy U can be expressed as: U = \frac{1}{2}kx^2 where x is the displacement from the equilibrium position.

Step 3: Introduce the given constraint of half-maximum kinetic energy

We are given that K is half of its maximum value. The maximum kinetic energy can be expressed as: K_max = \frac{1}{2}kA^2 where A is the maximum displacement (amplitude). Since K is half of K_max: K = \frac{1}{2}K_max = \frac{1}{4}kA^2

Step 4: Substitute the expressions for K and U in the equation for E

We have E as K + U, with K and U defined in steps 2 and 3. Substituting the expressions for K and U, we get: E = \frac{1}{4}kA^2 + \frac{1}{2}kx^2

Step 5: Express the maximum displacement A in terms of total energy E

Applied force at maximum displacement is given by Hooke's law: F = -kA Energy at maximum displacement is given by potential energy U, since there is no kinetic energy at that point: E = \frac{1}{2}kA^2 Solving for A, we get: A^2 = \frac{2E}{k}

Step 6: Substitute A^2 back into the equation from step 4, and solve for x

Replacing A^2 in the equation from step 4, we get: E = \frac{1}{4}k (\frac{2E}{k}) + \frac{1}{2}kx^2 => E = \frac{1}{2}E + \frac{1}{2}kx^2 Now, solving for x^2: x^2 = \frac{1}{k} (E - \frac{1}{2}E) = \frac{1}{2} \frac{2E}{k} = \frac{1}{2}A^2

Step 7: Express the displacement as a fraction of maximum displacement and find its square root

Now that we have x^2 in terms of A^2, we can express x as a fraction of A: \frac{x}{A} = \sqrt{\frac{x^2}{A^2}} = \sqrt{\frac{1}{2}} So, when the mass has half of its maximum kinetic energy, it is: x = \sqrt{\frac{1}{2}}A away from its equilibrium position expressed as a fraction of its maximum displacement A. This means that the mass is approximately 0.707 times the maximum displacement away from the equilibrium position when it has half of its maximum kinetic energy.