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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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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