Question

Il A block attached to a horizontal spring is pulled back a certain distance from equilibrium, then released from rest at 0 s. If the frequency of the block is 0.72 Hz, what is the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy?

Answer 1

The earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy is determined by the characteristics of the oscillating system.

Step-by-step explanation:

In this scenario, we have a block attached to a horizontal spring that is pulled back and then released. We are asked to find the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy.

Based on the given information, the frequency of the block is 0.72 Hz. When the block is at its earliest time, the velocity and kinetic energy are equal to zero. At this point, the force on the block is equal to the positive value of the spring constant multiplied by the amplitude, and the potential energy stored in the spring is equal to the spring constant multiplied by the square of the amplitude.

The total energy of the system, which is the sum of the potential energy and the kinetic energy, remains constant throughout the oscillations. Therefore, at the point where the kinetic energy is one-half of the potential energy, the total energy is also unchanged.

Answer 2

In simple harmonic motion, the total energy of the system is constant and equal to the sum of the potential energy and the kinetic energy. We can calculate the time when the block's kinetic energy is one-half of its potential energy by finding the position where the velocity and kinetic energy are equal to zero.

Step-by-step explanation:

The problem states that the block attached to a horizontal spring is pulled back from equilibrium and then released from rest at 0 s. The frequency of the block is given as 0.72 Hz. We need to find the earliest time after the block is released that its kinetic energy is exactly one-half of its potential energy.

In simple harmonic motion (SHM), the total energy of the system is constant and equal to the sum of the potential energy and the kinetic energy. The potential energy stored in the spring is given by U = (1/2)kA², where k is the spring constant and A is the amplitude. The kinetic energy is given by K = (1/2)mv², where m is the mass of the block and v is its velocity.

Since the frequency of the block is given, we can calculate the period T using the formula T = 1/f. Once we have the period, we can determine the time t when the kinetic energy is one-half of the potential energy. At this time, the block will be at a position x = -A, where the velocity and kinetic energy are equal to zero.