The motion of a particle connected to a spring is described by x=10cos(πt). At what time (in s) is the potential energy equal to the kinetic energy?

Question

The motion of a particle connected to a spring is described by x=10cos(πt). At what time (in s) is the potential energy equal to the kinetic energy?

Answer 1

That would be when the absolute value of the deflection is sqrt(1/2) of the maximum value. At those times, X^2 is half the maximum value, and so is the potential energy.

That happens when pi*t = pi/4 or 3*pi/4 or 5 pi/4 or 7 pi/4.
t = 1/4, 3/4, 5/4 or 7/4 s.

In this case the period is 2 seconds.

Answer 2

Thank you so much!

Answer 3

To find the time when the potential energy equals the kinetic energy, we need to set the equations for potential energy and kinetic energy equal to each other.

The potential energy of a particle connected to a spring can be given by the equation:
Potential Energy (PE) = 1/2 k x^2

The kinetic energy of a particle can be given by the equation:
Kinetic Energy (KE) = 1/2 m v^2

In this case, since the motion of the particle is described by x=10cos(πt), we can find the velocity by taking the derivative of the displacement equation with respect to time.

Given:
x = 10cos(πt)

Differentiating with respect to time, we get:
v = dx/dt = -10πsin(πt)

Now, let's substitute the expressions for potential energy (PE) and kinetic energy (KE) into an equation and solve for the time (t) when they are equal:

1/2 k x^2 = 1/2 m v^2

Using the given value of the displacement x=10cos(πt) and the velocity v=-10πsin(πt), we have:

1/2 k (10cos(πt))^2 = 1/2 m (-10πsin(πt))^2

Simplifying further,

k (100cos^2(πt)) = m (100π^2sin^2(πt))

Dividing both sides by 100,

k cos^2(πt) = m π^2sin^2(πt)

Now we can simplify the equation further:

k (1 - sin^2(πt)) = m π^2sin^2(πt)

k - k sin^2(πt) = m π^2sin^2(πt)

Rearranging the terms, we have:

m π^2sin^2(πt) + k sin^2(πt) = k

( m π^2 + k ) sin^2(πt) = k

sin^2(πt) = k / ( m π^2 + k )

Now, we can solve for t by taking the inverse sine (arcsin) of both sides:

πt = arcsin( √(k / ( m π^2 + k )) )

t = ( arcsin( √(k / ( m π^2 + k )) ) ) / π

With the given values of k and m, you can substitute those into the equation to get the time (t) in seconds when the potential energy equals the kinetic energy.

Answer 4

To find the time at which the potential energy is equal to the kinetic energy, we need to set up and solve the equation for potential energy and kinetic energy, and then find when they are equal.

The potential energy (PE) of a particle connected to a spring is given by the equation:

PE = (1/2)kx^2

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

The kinetic energy (KE) of a particle is given by the equation:

KE = (1/2)mv^2

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

Given the equation x = 10cos(πt), we can find the velocity by taking the derivative of x with respect to time:

v = dx/dt = -10πsin(πt)

Now, let's substitute the expressions for x and v into the equations for potential energy and kinetic energy:

PE = (1/2)k(10cos(πt))^2
KE = (1/2)m(-10πsin(πt))^2

To find the time at which PE = KE, we set up the equation:

(1/2)k(10cos(πt))^2 = (1/2)m(-10πsin(πt))^2

Simplifying this equation, we have:

k(10cos(πt))^2 = m(-10πsin(πt))^2

Divide both sides by 100, and divide by -1 on the right side to simplify further:

kcos^2(πt) = mπ^2sin^2(πt)

Now, we need to use trigonometric identities to simplify the equation. The identity we'll use is:

cos^2(θ) = 1 - sin^2(θ)

Substituting this identity into the equation, we have:

k(1 - sin^2(πt)) = mπ^2sin^2(πt)

Distribute k on the left side:

k - ksin^2(πt) = mπ^2sin^2(πt)

Rearranging the terms:

k = (mπ^2sin^2(πt)) + (ksin^2(πt))

Now, let's solve this equation for t.

First, we can simplify the equation by factoring out sin^2(πt):

k = sin^2(πt)(mπ^2 + k)

Divide both sides of the equation by (mπ^2 + k):

sin^2(πt) = k / (mπ^2 + k)

Applying the trigonometric identity:

sin^2(πt) = k / (mπ^2 + k) = 1 - cos^2(πt)

Now, we can set the equations for sin^2(πt) equal to each other and solve for t:

1 - cos^2(πt) = k / (mπ^2 + k)

Since cos^2(πt) + sin^2(πt) = 1, we can substitute 1 - cos^2(πt) with sin^2(πt):

sin^2(πt) = k / (mπ^2 + k)

Now, we can take the square root of both sides to isolate sin(πt):

sin(πt) = √[k / (mπ^2 + k)]

To solve for t, we need to find the inverse sine function of both sides:

πt = arcsin(√[k / (mπ^2 + k)])

Finally, divide both sides by π to solve for t:

t = arcsin(√[k / (mπ^2 + k)]) / π

This equation gives us the time (t) at which the potential energy is equal to the kinetic energy for the given particle connected to a spring, described by the equation x = 10cos(πt).