Find an expression for the impulse imparted by a force \(F(t)=F_{0} \sin (a t)\) during the time \(t=0\) to \(t=\pi / a .\) Here \(a\) is a constant with units of \(\mathrm{s}^{-1}\).

Short Answer

Expert verified
The impulse imparted by the force during the time \(t=0\) to \(t=\pi / a\) is given by \(2F_{0}/a\).

Step by step solution

01

Write down the expression for impulse

Impulse is given by the integral of the force with respect to time, symbolized as \( I = \int_{t_1}^{t_2} F(t) dt \). This equation reflects the mathematical interpretation of impulse as the area under a force-time curve.
02

Substitute the expression for force into the integral

With force being a sine function, \( F(t) = F_{0} \sin (at) \), we can rewrite the impulse integral as \( I = \int_{0}^{\pi/a} F_{0} \sin(at) dt \) .
03

Integrate the function

To integrate a sine function, we use the integral of sine, which is \(-\cos\). The impulse integral will be: \( I = -\frac{F_{0}}{a} [\cos(at)]_{0}^{\pi/a} \)
04

Substitute the limits into the function and simplify

We substitute the limits of the integral into the equation from step 3: \( I = -\frac{F_{0}}{a} [\cos(\pi) - \cos(0) ] \). Simplification gives \( I = -\frac{F_{0}}{a} [-1 -1] = \frac{2F_{0}}{a} \) .

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