A 110 g block connected to a light spring with a force constant of k = 6N/m osci

Question

A 110 g block connected to a light spring with a force constant of k = 6N/m oscillates on a horizontal, frictionless surface. The block is released from rest with an initial displacement xi = 5cm from equilibrium.

(B) At what value of x is the force from the spring equal to 76% of the maximum force?

Set | F | = | -kx | equal to 76% of the maximum value, without keeping track of the minus sign because the question asks about the magnitude of the force:

From this it is seen that the force is reduced to 76% of its maximum magnitude where |x| has 76% of its maximum value, or at x = ± ____cm

(C) At what value of x is the speed of the block equal to 76% of its maximum speed? We can answer this question from the solutions for the position and velocity as functions of time. Because the block is at its maximum displacement x = x0 at time t = 0, the equation for x is:

(1) x = x0 cos(t)

with = 0. Therefore the equation for the velocity is:

(2) v = -x0 sin(t)

whose maximum magnitude is x0. The time t when the speed is 76% of this maximum value therefore satisfies:

This is satisfied for:

But because sin2(t) + cos2(t) =1, Equation (4) reduces to:

And therefore the locations were the speed has the required values are:

_____cm

Explanation / Answer

B)

x = 0.76*5 cm = 3.8 cm

C)

x = 5 cm * sqrt(1-(0.76)^2)

x = ± 3.24 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.