# 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...

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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### 1 Answer

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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