What is the amplitude of the subsequent oscillations, What is the block's speed

Question

A 0.750 kg block is attached to a spring with spring constant 18 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 31 cm/s.

What is the amplitude of the subsequent oscillations?answer in cm and in two significant figures

What is the block's speed at the point where x = 0.50 A?Answer in cm/s and in two significant figures

Answer 1

I have no bloody clue as to how you would go about this without differential equations.  If you know calculus, this will answer you, if not... I'm sorry =)

Hooke's Law states:

F=-kx

which is a differential equation;

m d^2x/dt^2 = -kx

d^2x/dt^2 = -k/m x

let, w^2=k/m

d^2x/dt^2 = -w^2 x

which has a general solution

x(t) = A sin(w*t) + B cos(w*t)

However, we have two initial conditions,

I) x(0)=0             II) d/dt x(t)=v      where v=31 cm/s or .31 m/s

In order to satisfy the first initial condition B=0

So, we get the equation,

x(t)=A sin(w*t)

Here A is the amplitude of the oscillation.

To satisfy the second initial condition, we need to take a time derivative.

d/dt x(t) = w*A cos(w*t)

We want this to equal the initial velocity of 31 cm/s or .31 m/s at time t=0

v=w*A cos(0)

v= w*A

A=v/w

A=v/sqrt(k/m),  where sqrt() denotes the square root of the term inside the parenthesis.

A=.31m/s /sqrt(18(Nm^-1)/.75kg)

A=0.06327 m  or   A=6.33 cm

Hope this helps

Answer 2

2.     k = M a / A

18=(.750 Kg ) (x) / (.50)

divide this    do both sides you get       12 m/s     = 1,200 cm/s  

for number one i have no idea.