Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems - Page 405: 61

Answer

a) $2\;\rm Hz$ b) $1.2\;\rm cm$

Work Step by Step

a) We have here a system that undergoes a simple harmonic motion. And we know that the frequency of such a system is independent of the amplitude and the maximum velocity since it is given by $$\omega=2\pi f=\sqrt{\dfrac{k}{m}}$$ $$f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}$$ We can see that the right side is constant, so the frequency remains the same after the force blowing is finished. $$f'=f=\color{red}{\bf 2}\;\rm Hz$$ --- b) Before applying the thrust force, the maximum speed of the block is given by the conservation of energy. $$\color{red}{\bf\not} \frac{1}{2} mv_{1,\rm max}^2=\color{red}{\bf\not} \frac{1}{2} kA_1^2$$ So, $$mv_{1,\rm max}^2= kA_1^2$$ Noting that we used the maximum velocity since the force thrust applied on the block when it passes through the equilibrium point at which the speed is maximum. Thus, $$ v_{1,\rm max}^2 =\dfrac{ k}{m}A_1^2$$ Taking the square root, $$ v_{1,\rm max} =\sqrt{\dfrac{ k}{m}}A_1 $$ where $\sqrt{\dfrac{ k}{m}}=\omega=2\pi f$ $$ v_{1,\rm max} =2\pi f A_1 \tag 1$$ Noting that $\omega$ and $f$ remain constant before and after applying the thrust force. We know that the force provides an impulse on the block, so $$J_x=\Delta p=F\Delta t$$ Hence, $$mv_{2,\rm max}-mv_{2,\rm max}=F\Delta t$$ Plugging from (1); $$mv_{2,\rm max}-m(2\pi f A_1)=F\Delta t$$ We know that the velocity of the block at the equilibrium point is given by $A\omega$, so $v_{2,\rm max}=\omega A_2=2\pi f A_2$. $$2\pi f A_2m - 2\pi f A_1m =F\Delta t$$ Solving for $A_2$; $$ A_2 =\dfrac{F\Delta t+2\pi f A_1m}{2\pi f m} =\dfrac{F\Delta t}{2\pi f m}+ A_1 $$ Plugging the known, and recall that the force direction is toward the left while the block's initial velocity was to the right. $$ A_2 =\dfrac{(-20) (1\times 10^{-3})}{2\pi (2) (0.2)}+ 0.02=\bf 0.0120\;\rm m$$ $$A_2 =\color{red}{\bf 1.20}\;\rm cm$$
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