Physics ch.3 #141 help

Question

A 0.800 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 33 cm/s.

What is the amplitude of the subsequent oscillations?

What is the block's speed at the point where x= 0.40 A?

Answer 1

Let:

v be the speed immediately after the blow,

w be the angular frequency of the SHM,

a be the amplitude,

k be the spring constant,

m be the mass.

After the blow:

v = wa ...(1)

The angular frequency of the SHM satisfies:

w^2 = k / m ...(2)

Eliminating w:

v^2 / a^2 = k / m

a^2 = mv^2 / k

a = v sqrt(m / k) ...(2)

= 0.33 sqrt(0.800 / 15)

= 0.0762 m

= 7.62 cm.

If v1 is the speed when x = 0.40a:

v1 = w sqrt(a^2 - 0.40^2 a^2)

= wa sqrt(1 - 0.40^2)

Substituting for wa from (1):

v1^2 = v^2(1 - 0.40^2)

v = 0.33 sqrt(1 - 0.40^2)

= 0.302 m/s.

= 30.2 cm/s.