Distance for Nuclear Shockwave to Destroy a House

[vladimir69]

Shockwave destroying a house [Solved]

Homework Statement

A 1-megaton nuclear explosion produces a shock wave whose amplitude, measured as excess air pressure above normal atmospheric pressure, is 1.4x10^5 Pa at a distance of 1.3km from the explosion. An excess pressure of 3.5x10^4 Pa will destroy a typical wood-frame house. At what distance from the explosion will such houses be destroyed? Assume the wavefront is spherical.

Homework Equations

$$I=\frac{P}{A}$$
$$A=4\pi R^2$$

The Attempt at a Solution

I just treated the excess air pressure as "intensity". Assuming the force drops off as the inverse square of distance.
$$P=IA=1.4\times10^5\times 4 \pi \times 1.3^2=2973203.3$$
$$P=I_1 A_1$$
$$2973203.3=3.5\times 10^4 \times 4\pi R^2$$
$$R^2=6.76km^2$$
$$R=2.6km$$
in the book the answer gives 5.6km, looks like i am out by a factor of 2, not sure how though.

 

[ideasrule]

But amplitude is not intensity; rather, intensity is proportional to the square of amplitude. Think about the simplest oscilliating system: a spring. The energy stored is kA^2, where A is amplitude. Since intensity is simply energy per time divided by area, and since the shockwave is an oscillation, intensity scales as the square of the amplitude as well.

 

[vladimir69]

Thanks mate, got it!