A mass that weighs 8 lb stretches a spring 24 in. The system is acted on by an external force of 4 sin 4t lb. If the mass is pulled down 6 in. and then released, determine the position of the mass at any time. Determine the first four times at which the velocity of the mass is zero

Answers

Answer 1
Answer:

Answer:

t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)

Explanation:

The equation of force is

F = 4 sin 4 t

Compare with the standard equation

f = A sin wt

where, w is the angular frequency and A is the amplitude.

Now

w = 4 rad/s

Let the time period is T.

the relation for the time period is

T = (2\pi)/(w)\n\nT = (2 \pi)/(4)\n\nT = (\pi)/(2)

the time period is defined as the time taken by the body to complete one oscillation.

So, the velocity is zero at the extreme points where the object is at time, T/4 and its odd T/2, 3T/4, 3T/2, etc.

So, the velocity is zero at time

t = (\pi)/(8), (\pi)/(4), (3\pi)/(8), (3 \pi)/(4)

Answer 2
Answer:

Final answer:

To determine the position of the mass and the times when the velocity is zero in a mass-spring system.

Explanation:

To determine the position of the mass at any time, we need to solve the equation of motion for the mass-spring system. The equation is:

mx'' + kx = Fext

where m is the mass, x is the displacement of the mass from its equilibrium position, k is the spring constant, and Fext is the external force. We can solve this differential equation to find the position of the mass as a function of time.

To determine the first four times at which the velocity of the mass is zero, we need to solve for the velocity of the mass, which is given by:

v = x'

We can find the times at which the velocity is zero by finding the values of t for which x' = 0.

Learn more about Mass-spring system here:

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A. x>0 but xB. x=A
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Answers

Answer:

The correct answer is option E. This is because when a body is undergoing simple harmonic motion, the displacement, velocity and acceleration are constantly changing with respect to position and time.

Explanation:

When the body is passing through the equilibrium position so that the displacement at that point is zero, the velocity at that point is either maximum or minimum (Vmax or Vmin) which is dependent on which way the body is moving (the chosen direction of positive sense of motion). At the equilibrium position the acceleration is zero because the net force acting on the body is zero (that is the sun of the forces acting on the body along both directions of motion is equal to zero). Also when the body is at either its maximum positive displacemt, x = +A, or its maximum negative displacement x = -A, the velocity is zero and the body is instantaneously at rest. At this points the force (in either direction) acting on the body has it maximum value and as a result the acceleration has it maximum magnitude whether negative or positive depending on the chosen direction positive sense of motion.

At any other point in the course of the SHM (simple harmonic motion), the velocity (V) and acceleration (a) of the always have nonzero values and as a result they are never simultaneously equal to zero.

Thank you very much reading and I hope this is helpful to you.

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Answers

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