Monday

January 26, 2015

January 26, 2015

Posted by **Susan** on Friday, December 3, 2010 at 1:55pm.

A 45 g mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:

y(t) = 1.4 * sin( 4 * t ) where y is measured in meters and t in seconds.

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(a) What is the spring constant in N/m?

k = N/m *

0.72 OK

HELP: You are given m, the mass. What other quantity appears in the equation involving k, the spring constant, and m?

HELP: You are given the equation of motion

y(t) = A * sin( ù * t )

Now can you find the missing quantity?

HELP: Be careful of the units of m.

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(b) What is the Total Energy in the mass and spring in J?

E = J

HELP: The total energy is the sum of the kinetic energy and potential energy. At what point in the motion is the energy all kinetic? At what point is it all potential? Can you compute it at this point?

HELP: Kinetic energy is 1/2 * Mass * v2

Potential energy is 1/2 * k * y2

We know that at the mass's maximum y-position it has zero velocity, so simply compute Potential Energy at that point.

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(c) What is the maximum Kinetic Energy of the mass?

KE = J

HELP: Remember that Total Energy = Kinetic + Potential Energy.

HELP: When the mass is at y=0, it has zero potential energy. So what must its kinetic energy be at that point?

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(d) What is the maximum velocity of the mass in m/s ?

v(max) = m/s

HELP: Maximum velocity occurs at maximum kinetic energy.

HELP: KE = 1/2 * Mass * v2

Solve for v using the result from part (c).

- Physics -
**Kristin**, Sunday, December 5, 2010 at 11:30pmfor part b, take the amplitude (1.4) and square it and then multiply by what you got for part a.

so PE = .5 * k * x^2

= .5 * .72 * 1.4^2

- Physics -
**Kristin**, Sunday, December 5, 2010 at 11:32pmfor part c, the answer should be the exact same as part b.

- Physics -
**Kristin**, Sunday, December 5, 2010 at 11:35pmand for part d, take what the answer you got for c and plug all your variables into the equation of

answer part c = .5 * .045 * v^2

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