Acoustics Problem #1

The steady-state response of a simple dynamic system to a sinusoidal excitation force is…

$F(t)=2.248\sin (0.1\pi t){\mathrm{lb}}_{\mathrm{f}}$

and the resulting displacement response is:

$x(t)=0.328\sin (0.1\pi t-30°)\mathrm{ft}$

where the displacement lags the force by a phase angle $\varphi =30°$.

Determine

1. the work done by the excitation force in one minute
2. the work done by the excitation force in one second

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Solution

Given

• $F_0=2.248{\mathrm{lb}}_{\mathrm{f}}$
• $A=\mathrm{0.328\; ft}$
• $\omega =0.1\mathrm{\pi \; rad/sec}$
• $\varphi =30°$

Work done per cycle

For harmonic excitation, the work done per cycle is $W_{\text{cycle}}=\pi A{F}_0\sin \varphi$.

Substitute $A=\mathrm{0.328\; ft}$, $F_0=2.248{\mathrm{lb}}_{\mathrm{f}}$, and $\sin \varphi =\sin \mathrm{(30°)}=\mathrm{½}$:

$W_{\text{cycle}}= \pi (0.328)(2.248)(½)=\mathrm{1.159\; ft\cdot lbf/cycle}$

Period

$T=\frac{2\pi }{\omega }=\frac{2\pi }{0.1\pi }=20\text{\hspace{0.17em}sec}$

Cycles in one minute: $N=\frac{60}{20}=3$

(a) Work done in one minute

$W_{\text{60s}}=N{W}_{\text{cycle}}= =\mathrm{3.48\; ft\cdot lbf}$

(b) Work done in one second

Work over an interval is $W={\int }_0^{1}F(t)x(t)dt$, where $ẋ(t)=A\omega \cos (\omega t-\varphi )$.

So, $W_{\text{1s}}={\int }_0^1\left[F_0\sin (\omega t)\right][A\omega \cos (\omega t-\varphi )]dt$

Evaluating the integral for $F_0=2.248$, $A=0.328$, $\omega =0.1\pi$, $\varphi =30°$ gives:

$W_{\text{1s}}=\mathrm{0.0342\; ft\cdot lbf}$

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Final Answers

• (a) $W_{\text{60s}}=\mathrm{3.48\; ft\cdot lbf}$
• (b) $W_{\text{1s}}=\mathrm{0.0342\; ft\cdot lbf}$