Question

A 0.700kg block is attached to a spring with spring constant 16N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 31cm/s . What is the amplitude of the subsequent oscilllations? What is the block's speed at the point where x= 0.30

Answer 1

Assume that there is no lost in energy.
So, the kinetic energy given to the block will be changed into the potential energy when the block is at its maximum displacement, i.e. amplitude. Hence,
1. kinetic energy = potential energy stored in the spring
<=> 0.5mu^2 = 0.5kA^2
<=> mu^2 = kA^2
<=> A^2 = mu^2/k
<=> A = sqrt(m/k) u = sqrt(0.7/18)(31) = 6.1132cm

2.Total energy=kinetic energy + potential energy in the spring = 0.5mv^2 + 0.5kx^2 , where x is the position of the block. Total energy is just the initial kinetic energy,E
= 0.5mu^2=0.033635J
Hence, E = 0.5mv^2 + 0.5kx^2
<=>0.5mv^2 = E - 0.5kx^2
<=>v^2=2E/m - kx^2/m
<=>v=sqrt(2E/m - kx^2/m)=sqrt(2x0.033635/0.7 - 18(0.03)^2/0.7)
=sqrt (0.09) = 0.27ms^-1 = 27cms^1 #