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A block attached to a horizontal spring of force constant 75N/m undergoes SHM with an amplitude of 0.15m. If the speed of the mass is 1.7 m/s when the displacement is 0.12m from the equilibrium position, what is the mass of the block?

Respuesta :

Answer:

x = A sin w t         for SHM where w = angular frequency

sin w t = x / A = .12 / .15 = .8    where w t is the angle in degrees

w t = 53.1 deg       (w itself is in rad / sec)

since v = A w cos w t  

then v / x = w cos w t / sin w t = w / tan w t

w = v / x * tan 53.1 = 1.7 / .12 * tan 53.1 = 18.9 /sec

Also for SHM    w = (k / m)^1/2

m = k / w^2 = 75 / 18.9^2 = .21 kg

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