Physics- a mechanics question

2012-10-06 9:12 pm

A 0.5 kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm longer than its equilibrium length, the speed of the block is 0.5 m/s. What is the greatest speed of the block?

回答 (1)

天同

2012-10-07 3:39 am

✔ 最佳答案

The angular frequency w of a spring-mass system is given by,
w^2 = k/m
where k is the spring constant (= 80 N/m)
m is the mass of the block (=0.5 kg)
hence, w^2 = 80/0.5 s^-2 = 160 s^-2
w = 12.65 s^-1

The speed of the oscillating block v is,
v = w.square-root[A^2 - x^2]
where A is the amplitude of oscillation
x is the displacement from the equilibrium position
hence, 0.5 = 12.65[square-root(A^2 - 0.04^2)]
A = 0.05623 m

The max speed occurs at the equilibrium position, i.e. x = 0 m
v(max) = w.A = 12.65 x 0.05623 m/s = 0.711 m/s

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