Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is 1/8 of its amplitude. (The answer is an integer.)

Approach: Choose a specific trigonometric form for the position function x(t). It doesn't matter which since the answer doesn't depend on the initial conditions. Let the amplitude be 'A' (which will cancel when you compute the ratio). Determine v(t) that corresponds to your choice of x(t). Apply the given displacement condition to determine what your choice of x(t) was equal to (you don't need to try and compute a time: the result holds for any simple harmonic oscillator). From this directly compute the desired ratio. The trig identity sin2(è)+cos2(è)=1, which as discussed in class is equivalent to the conservation of energy here, will definitely be helpful.

P.E. is proportional to the square of displacement. When displacement is 1/8 of max, the P.E. is 1/64 of the maximum value, which is also the total energy. The rest (63/64 of total energy) is kinetic energy.

The KE/PE ratio at the time in question is therefore 63.

Thanks for the help!

To solve this problem, we need to use the definition of kinetic energy and potential energy of a simple harmonic oscillator.

The kinetic energy (KE) of a simple harmonic oscillator is given by the formula:

KE = (1/2) * m * v^2

where m is the mass of the oscillator and v is its velocity.

The potential energy (PE) of a simple harmonic oscillator is given by the formula:

PE = (1/2) * k * x^2

where k is the spring constant and x is the displacement of the oscillator from its equilibrium position.

First, let's choose a specific trigonometric form for the position function x(t). Since the answer doesn't depend on the initial conditions, we can choose any form that represents a simple harmonic oscillator. Let's use the form:

x(t) = A * sin(ωt)

where A is the amplitude of the oscillator and ω is the angular frequency.

Now, let's find the velocity function v(t) that corresponds to x(t). Differentiating x(t) with respect to time, we get:

v(t) = A * ω * cos(ωt)

Next, we will apply the given displacement condition, which states that the displacement is 1/8 of the amplitude. So, we have:

x(t) = 1/8 * A

Substituting the chosen form of x(t), we get:

A * sin(ωt) = 1/8 * A

Dividing both sides by A, we obtain:

sin(ωt) = 1/8

To solve for ωt, we take the inverse sine of both sides:

ωt = sin^(-1)(1/8)

Using a calculator, we find ωt ≈ 0.1244 radians.

Now, using the trig identity sin^2(è) + cos^2(è) = 1, we can calculate the cosine of ωt:

cos(ωt) = √(1 - sin^2(ωt))

cos(ωt) = √(1 - (1/8)^2)

cos(ωt) ≈ √(1 - 1/64) ≈ √(63/64) ≈ 0.9921

Finally, substituting our results into the formulas for KE and PE, we get:

KE = (1/2) * m * (A * ω * cos(ωt))^2 = (1/2) * m * A^2 * ω^2 * cos^2(ωt) = (1/2) * m * A^2 * ω^2 * 0.9921^2

PE = (1/2) * k * (1/8 * A)^2 = (1/2) * k * (1/64) * A^2

The ratio of kinetic energy to potential energy is given by:

KE/PE = [(1/2) * m * A^2 * ω^2 * 0.9921^2] / [(1/2) * k * (1/64) * A^2]

The A^2 terms cancel out, and we can simplify the rest of the expression:

KE/PE = (m * ω^2 * 0.9921^2) / (k/64)

Since the answer should be an integer, we can choose suitable values for m, ω, and k to make the expression an integer. For example, let's choose m = 1 kg, ω = 1 rad/s, and k = 1 N/m.

Substituting these values, we get:

KE/PE = (1 * (1)^2 * 0.9921^2) / (1/64) = (0.9921^2) / (1/64) = 0.9921^2 * 64

Evaluating this expression, we find:

KE/PE ≈ 39.88

Therefore, the ratio of kinetic energy to potential energy of the simple harmonic oscillator is approximately 39.88.