A 0.58 kg mass is attached to a light spring with a force constant of 38.9 N/m a

Question

A 0.58 kg mass is attached to a light spring with a force constant of 38.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following.

(a) maximum speed of the oscillating mass
________ m/s

(b) speed of the oscillating mass when the spring is compressed 1.5 cm
________ m/s

(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
_________ m/s

(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value
_________ m

Explanation / Answer

x= A cos wt--equation for mass undergoing SHM

x= 5 cos (wt)

k = 38.9 N/m, m = 0.58 kg , w= sqroot ( k/m) = 8.2

x= (0.05) cost ( 8.2t)

a) Max Speed = Aw= 0.41

b)max KE = 1/2 mv^2 = 0.5 ( 0.58)( 0.41^2)

at comp[ression of 1.5 cm

0.5 ( 0.58) ( 0.41^2) = 0.5 ( 0.58)( v^2) + 1/2 ( 38.9) ( 0.000225)

v = 0.39 m/s

c) we will use conservation of energy to derive the expression for velocity

1/2kA^2 = 1/2 mv^2 + 1/2 kx^2 ( where A is the amplitude)

kA^2 - kx^2 = mv^2

v^2= k/m ( A^2- x^2)

v= w sqroot ( A^2- x^2)

v = 8.2 saroot ( 0.0025 - 0.000225) = 0.39 m/s

d) w sroot ( A^2- x^2) = 1/2 ( w) A

sqroot ( 0.0025 - x^2) = 0.5 ( 0.05)

0.0025 - x^2= 0.000625

0.0025 - 0.000625 = x^2

x= 4.33 cm

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