Question

A certain object of mass 0.50 kg moves along the x-axis. Its potential energy is given by U = 3x2 + 8x. Starting at rest, the object is released at the origin. (a) Find the maximum speed attained by the object. (b) Find the position about which it oscillates. (c) Find the frequency of the oscillation. 20. a. 4.6 m/s b. x = −1.3 m c. 0.551 Hz

Answer 1

The object reaches its maximum speed when all the potential energy is converted into kinetic energy. The maximum speed can be found by determining the position where the potential energy is minimized and using the conservation of mechanical energy. The frequency of the oscillation can be found using the formula for the frequency of simple harmonic motion.

Step-by-step explanation:

To find the maximum speed attained by the object, we need to determine the position where the potential energy is minimized and the kinetic energy is maximized. The object reaches its maximum speed when all the potential energy is converted into kinetic energy. This happens when the object is at a position where the slope of the potential energy curve is zero, i.e., the derivative of the potential energy with respect to position is zero. Therefore, we need to find the position where the derivative of the potential energy equation, U, is zero.

We have U = 3x^2 + 8x. Taking the derivative of U with respect to x, we get dU/dx = 6x + 8. To find the position where dU/dx is zero, we set it equal to zero and solve for x:

6x + 8 = 0 -> 6x = -8 -> x = -8/6 = -1.33 m

So, the object oscillates around the position x = -1.33 m. To find the maximum speed, we first need to find the maximum potential energy, which occurs at the equilibrium position. And then, we can find the maximum speed using the conservation of mechanical energy.

At the equilibrium position, the potential energy is given by U = 3x^2 + 8x:

U = 3(-1.33)^2 + 8(-1.33) = -2.99 J

Since the total mechanical energy is given as -6 J, we can find the maximum kinetic energy using E = U + K:

-6 J = -2.99 J + K -> K = -3.01 J

The maximum kinetic energy is equal to the maximum potential energy, which occurs at the equilibrium position. Using the equation for kinetic energy, K = (1/2)mv^2, we can solve for the maximum speed:

-3.01 J = (1/2)(0.5 kg)v^2 -> v^2 = -6.02 J/kg -> v = sqrt(-6.02 J/kg) ≈ 2.45 m/s

Therefore, the maximum speed attained by the object is approximately 2.45 m/s.

The frequency of the oscillation can be found using the formula for the frequency of simple harmonic motion, f = (1/2π) * sqrt(k/m), where k is the spring constant and m is the mass of the object. In this case, the spring constant is 8 N/m and the mass is 0.5 kg:

f = (1/2π) * sqrt(8 N/m / 0.5 kg) = (1/2π) * sqrt(16) ≈ 0.551 Hz

Therefore, the frequency of the oscillation is approximately 0.551 Hz.

Answer 2

To determine the properties of an object in SHM, conservation of energy is applied to find the maximum speed, differentiation of potential energy is used for finding the equilibrium position, and the frequency of oscillation is calculated with the mass of the object and the effective force constant.

Step-by-step explanation:

To solve the problem of a 0.50 kg object undergoing oscillatory motion, we first need to understand the principles of conservation of energy and simple harmonic motion (SHM).

(a) The maximum speed attained by the object is when it has converted all its potential energy into kinetic energy. Using the given potential energy function U(x) = 3x2 + 8x and the conservation of energy, we can set the total mechanical energy (the sum of potential and kinetic energy) to zero because the object starts at rest from the origin. We find the position where the potential energy is the least since this is where the kinetic energy, and therefore the speed, will be the greatest. Differentiating U(x) with respect to x and setting the derivative to zero gives the position of least potential energy, which allows us to calculate the maximum speed.

(b) The object oscillates about the position where the net force (and thus the derivative of the potential energy function) is zero. This is the equilibrium position, which can be found by differentiating U(x) and setting the result equal to zero.

(c) The frequency of oscillation can be found using the formula for the frequency of a SHM system, f = (1/2π) * √(k/m), where k is the force constant obtained from the second derivative of the potential energy function, and m is the mass of the object.