A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.076 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 17.8 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length?

x=Asinωt

v=dx/dt =Aω cosωt= Aω•sqrt(1-sin²ωt) =
=ω•sqrt(A² - A² sin²ωt)= ω•sqrt(A² -x²)
A=0.076 m
x= 0.045 m
ω =17.8 rad/s
v= ω•sqrt(A² -x²)= …

To find the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length, we can use the concept of energy conservation.

First, let's determine the spring constant, k, of the horizontal spring. The angular frequency, ω, and the spring constant, k, are related by the equation:

ω = √(k / m),

where m is the mass of the object attached to the spring.

Rearranging the equation, we have:

k = ω^2 * m.

Next, we can calculate the potential energy stored in the spring when it is stretched by 0.076 m. The potential energy of a spring is given by:

PE = 1/2 * k * x^2,

where x is the displacement from the equilibrium position.

Substituting the values into the equation, we have:

PE = 1/2 * k * (0.076)^2.

Then, we can calculate the potential energy of the spring when it is stretched by 0.045 m. Using the same formula:

PE' = 1/2 * k * (0.045)^2.

Since energy is conserved in the system without any losses, the change in potential energy is equal to the change in kinetic energy when the spring is stretched. Therefore:

PE - PE' = 1/2 * m * v^2,

where v is the speed of the object.

Substituting the values into the equation:

1/2 * k * (0.076)^2 - 1/2 * k * (0.045)^2 = 1/2 * m * v^2.

Now, we have everything we need to solve for v. First, let's calculate k:

k = (17.8 rad/s)^2 * m.

Then, substitute this value into the equation:

1/2 * [(17.8 rad/s)^2 * m] * (0.076)^2 - 1/2 * [(17.8 rad/s)^2 * m] * (0.045)^2 = 1/2 * m * v^2.

Simplifying the equation, we can solve for v:

[(17.8 rad/s)^2 * m] * [(0.076)^2 - (0.045)^2] = m * v^2.

[(17.8 rad/s)^2] * [(0.076)^2 - (0.045)^2] = v^2.

Calculate the value enclosed in square brackets:

[(17.8 rad/s)^2] * [(0.076)^2 - (0.045)^2] ≈ 3.356 rad^2/s^2.

Taking the square root of both sides, we find:

v ≈ √(3.356 rad^2/s^2).

Calculating the square root:

v ≈ 1.832 m/s.

Therefore, the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length is approximately 1.832 m/s.

To find the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length, we can use the formula for the speed of an object in simple harmonic motion.

The speed of the object in simple harmonic motion is given by the equation:

v = ω √(A^2 - x^2)

Where:
v = speed of the object
ω = angular frequency
A = amplitude of the motion
x = displacement of the object from the equilibrium position

Given:
ω = 17.8 rad/s
A = 0.076 m
x = 0.045 m

Substituting the given values into the formula:

v = 17.8 rad/s * √(0.076^2 - 0.045^2)

First, caculate (0.076^2 - 0.045^2):

v = 17.8 rad/s * √(0.005776 - 0.002025)

Next, find the square root:

v = 17.8 rad/s * √(0.003751)

v = 17.8 rad/s * 0.06123

Finally, calculate the product:

v ≈ 1.088 m/s

Therefore, the speed of the object at the instant when the spring is stretched by 0.045 m relative to its unstrained length is approximately 1.088 m/s.