The displacement of a block of mass 778g778g attached to a spring whose spring constant is 38N/m38N/m is given by x=Asin(ωt)x=Asin(ωt) where A=25cmA=25cm . In the first complete cycle find the values of xx and tt at which the kinetic energy is equal to one half the potential energy.

First position:
cm...... First time:
s.

Second position:
cm...... Second time:
s..

Third position:
cm...... Third time:
s.

Fourth position:
cm...... Fourth time:
s.

I already did one like this for you.

To find the values of x and t at which the kinetic energy is equal to one half the potential energy in the first complete cycle, we need to equate the equations for kinetic energy and potential energy.

The potential energy of the block attached to the spring can be given as:

PE = (1/2)kx^2

where k is the spring constant and x is the displacement of the block.

The kinetic energy of the block can be given as:

KE = (1/2)mv^2

where m is the mass of the block and v is the velocity of the block.

In this case, the potential energy is equal to the kinetic energy when:

(1/2)kx^2 = (1/2)mv^2

We can substitute the given values into this equation:

(1/2)(38 N/m)(x)^2 = (1/2)(0.778 kg)(v)^2

Simplifying the equation:

19x^2 = 0.778v^2

Since we are dealing with a harmonic motion, we know that the maximum velocity occurs when the displacement is zero (i.e., the block passes through the equilibrium position). Therefore, we can set v = 0 when x = 0.

Substituting these values into the equation:

19(0)^2 = 0.778(0)^2

0 = 0

This means that at the equilibrium position (x = 0), the kinetic energy is zero.

To find the values of x and t at which the kinetic energy is equal to one half the potential energy in the first complete cycle, we need to look at the amplitude of the motion (A).

In this case, we are given A = 25 cm. As we know, the amplitude represents the maximum displacement of the block from the equilibrium position. Therefore, the potential energy will be maximum at A and the kinetic energy will be zero. This occurs when the block is momentarily at rest before changing direction.

To find the time at which this occurs, we need to consider the equation x = A sin(ωt), where ω is the angular frequency. In this case, we are not given the value of ω, so we need to calculate it.

The angular frequency can be calculated as:

ω = 2πf

where f is the frequency of the motion. Frequency can be calculated as the reciprocal of the time period:

f = 1/T

where T is the time period, which represents the time taken for one complete cycle. In this case, we need to find the time period.

The time period can be calculated as:

T = 2π/ω

Given that x = A sin(ωt) represents the first complete cycle, we know that when x = A, sin(ωt) = 1.

Therefore, we can substitute these values into the equation:

A = A sin(2πfT)

1 = sin(2πfT)

To solve this equation, we can use inverse sine:

sin^(-1)(1) = 2πfT

π/2 = 2πfT

1/4 = fT

Since the frequency is the reciprocal of the time period, we can say:

1 = 4fT

Now we have an equation connecting the frequency and time period. However, we don't have the value of the time period or the frequency. Therefore, we need to consider the given information.

From the equation x = A sin(ωt), we are given A = 25 cm. This represents the maximum displacement and occurs when the block is momentarily at rest.

To find the time at which this happens, we need to evaluate the equation for x = A:

25 = 25 sin(ωt)

1 = sin(ωt)

This occurs when ωt = π/2.

To find the value of t, we can divide by ω:

ωt = π/2

t = (π/2) / ω

Remember that ω = 2πf. Substituting this back into the equation:

t = (π/2) / (2πf)

Simplifying:

t = (1/4f)

We can see that t is inversely proportional to the frequency.

Therefore, at the first position (x = A) when the block is momentarily at rest, the time can be calculated as t = (1/4f).

To calculate the other three positions and times, we need to consider the phase of the motion. The equation x = A sin(ωt) gives us the displacement of the block at any given time. By analyzing the equation, we can determine the positions and times at which the kinetic energy is equal to one-half the potential energy.

Without knowing the phase of the motion, it is not possible to determine the specific positions and times at which the kinetic energy is equal to one-half the potential energy in the first complete cycle.