Question

A $1.0kg$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 its a speed of $40cm/s$.

What are

a. The amplitude of the subsequent oscillations?

b. The block's speed at the point where $x=\frac{1}{2}A$?

Short answer

a) The amplitude of the subsequent oscillations is $A=10.0cm$

b) The block speed at the point when $x=\frac{1}{2}A$is $v=34.64cm/s$

Step-by-step solution

Given data.

Given:

• $m=1.0kg$
  
• $k=16N/m$
  
• $v\left(0\right)=40cm/s$
  
• $x\left(0\right)=0$
  

Required:

(a) A

(b)$v$when$x=\frac{1}{2}A$

Simplification.

a) using conservation law and simplify to get that

$v=\omega \sqrt{A^2-x^2}\Rightarrow A^2={\left(\frac{v}{\omega }\right)}^2+x^2\Rightarrow \left(1\right)$

All is known except $\omega$which is given by

${\omega }^2=\frac{k}{m}\Rightarrow \left(2\right)$

substitution in $\left(2\right)$to get that

${\omega }^2=\frac{16}{1}=16\Rightarrow \omega =4rad/s$

substitution in $\left(1\right)$to get that

$A^2={\left(\frac{0.4}{4}\right)}^2\Rightarrow A=0.1m=10.0cm$

Using equation (1)

To get

$v=4\sqrt{{\left(10\right)}^2-{\left(5\right)}^2}=34.64cm/s$

Where :

$x=A/2=10/2=5cm$