# Harmonic Oscillator. Gauge and Absolute Pressure

Question 1

A simple harmonic oscillator consists of a block of mass $3.30 kg$ attached to a spring of spring constant $210 N/m$. When $t = 1.70 s$, the position and velocity of the block are $x = 0.198 m$ and $v = 2.830 m/s$. What is the amplitude of the oscillations? What were the position and velocity of the block at $t = 0 s$?

The equation of displacement $X(t)$ of a harmonic oscillator is

$X(t) = A*\sin(omega*t+\phi)$ (1)

and thus the speed is

$V(t) = d X/d t = A*\omega*\cos(\omega*t+\phi)$ (2)

We also know that

$\omega = \sqrt{K/m} = \sqrt{210/3.3} =7.977 (rad/sec)$

by dividing (1) and (2) we obtain

$(1/\omega)*\tan(\omega*t+\phi) = (X/V)$

$\tan(\omega*t+\phi) = (X*\omega/V) = (0.198*7.977)/2.830 =0.558$

which is equivalent to

$\omega*t + \phi =0.509 rad$

At t = 1.7 sec

$\phi = 0.509-7.977*1.70 = -13.05 rad$

$A = X(1.7)/\sin(0.509) = 0.198/\sin(0.509) =0.406 m$

At t = 0 one has

$X(0) = 0.406*\sin(-13.05) =-0.189 m$

$V(0) = 0.406*7.977*\cos(-13.05) =2.87 m/s$

Question 2

A fire extinguisher contains a high-pressure liquid so that it can spray the liquid out very quickly when needed. If the fluid leaves a fire extinguisher at a speed of 37.9 m/s, what is the pressure inside? (Assume water is the liquid inside the fire extinguisher.)

The Bernoulli equation states that

$P = P_0+ (\rho*v^2)/2$

Is P is the ABSOLUTE pressure inside (NOT the gauge pressure) and $p0 =10^5 N/m^2$ the pressure outside, for $rho =1000 kg/m^3$ we obtain

$P = 10^5 +(1000*37.9^2)/2 = 8.182*10^5 (N/m^2) \simeq 8.18 (atm)$

Observation: The difference between the ABSOLUTE Pressure and GAUGE pressure is that the Gauge Pressure has already the Atmospheric pressure subtracted

from it, that is

$P(Gauge)= P(Absolute) – P_0$