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A block of unknown mass is attached to a spring with a spring constant of 10 N/m and...

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bobby9901 | Student, College Freshman | Salutatorian

Posted January 21, 2013 at 10:59 PM via web

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A block of unknown mass is attached to a spring with a spring constant of 10 N/m and undergoes simple harmonic motion with an amplitude of 8.0 cm.

 

When the block is 1/4 of the way between its equilibrium position and the endpoint, its speed is measured to be 30.0 cm/s.

calculate The mass of the block, the period of the motion.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted January 30, 2013 at 3:17 PM (Answer #1)

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The simple harmonic motion of a spring can be described by the following equations.

`x = Asinomegat = Asin(sqrt(k/m)t)`

`dotx^2 = (omega)^2(A^2-x^2)`

 

x = Displacement from the original position

omega = angular frequency

A = amplitude

k = spring constant

t = time

m = mass of the block

 

Using above we can get;

`omega = sqrt(k/m)`

 

`dotx^2 = (omega)^2(A^2-x^2)`

`dotx^2 = (k/m)(A^2-x^2)`

 

It is given that When the block is 1/4 of the way between its equilibrium position and the endpoint, its speed is measured to be 30.0 cm/s.

This means;

`dotx = 30`

`x = A/4 = (8)/4m = 2cm`

`k = 10N/m`

 

`dotx^2 = (k/m)(A^2-x^2)`

   `30^2 = (10/m)(8^2-2^2)`

       ` m = (10)(8^2-2^2)/(30^2) = 0.667kg`

 

So the mass of the block is 667g.

 

`omega = sqrt(k/m)`

`omega = sqrt(10/0.667)`

`omega = 3.872rad/s`

 

But `omega = (2pi)/T` where T is the the period of motion.

`omega = (2pi)/T`

        `T = (2pi)/omega`

        `T = (2pi)/(3.872)`

        `T = 1.623s`

 

So the period of oscillation is 1.623 seconds.

 

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