Question

A 0.900 kg block is attached to a spring with spring constant 14.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 34.0 cm/s . What are You may want to review (Pages 400 - 401) . Part APart complete The amplitude of the subsequent oscillations? Express your answer with the appropriate units. 8.47 cm Previous Answers Correct Here we learn how to calculate the amplitude of the spring's oscillations. Part B The block's speed at the point where x= 0.150 A? Express your answer with the appropriate units.

Answer 1

Answer:

1. $A=0.0847\ m$

2. $v=0.050\ m.s^{-1}$

Explanation:

Given:

• mass of block, $m=0.9\ kg$

• spring constant, $k=14.5\ N.m^{-1}$

• maximum velocity of block, $v_m=34\ cm.s^{-1}$

1.

We know from the energy of oscillating spring:

$E=\frac{1}{2}.k.A^2= \frac{1}{2}.m.v_m^2$

$\frac{1}{2}\times 14.5\times A^2=\frac{1}{2}\times 0.9\times 0.34^2$

$A=0.0847\ m$

2.

Now, given that:

instantaneous position, $x=0.15\times A=0.0127\ m$

we find angular speed,

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{14.5}{0.9} }$

$\omega=4.013\ rad.s^{-1}$

So we have

$v=x.\omega$

$v=0.0127\times 4.013$

$v=0.051\ m.s^{-1}$