Question #306588

A particle perform simple harmonic motion in a straight line about a center C in the line.When it's distances from C are 5.0 and 13.0cm,it's velocities are 25.0 and 7.0cm/s respectively.Find the period and the amplitude of the oscillation. Hint: use V=±w(angular velocity)√(A² - y²) and solve simultaneously.?

1
Expert's answer
2022-03-08T08:55:26-0500

Explanations & Calculations


  • Just like the hint says, apply that formula to create 2 simultaneous equations to solve for the needed variables.

25.0cms1=ωA2(5.0cm)225.0=ωA225(1)7.0=ωA2169(2)(1)÷(2)25.07.0=A225A2169A2=181.25A=13.5cmAin(1),25=ω181.2525ω=2rads1ω=2=2πTT=π=3.1s\qquad\qquad \begin{aligned} \small 25.0\,cms^{-1}&=\small \omega\sqrt{A^2-(5.0\,cm)^2}\\ \small 25.0&=\small \omega\sqrt{A^2-25}\cdots\cdots(1)\\ \\ \small 7.0&=\small \omega\sqrt{A^2-169}\cdots\cdots(2)\\ \\ \small (1)\div(2)\\ \\ \small \frac{25.0}{7.0}&=\small \sqrt{\frac{A^2-25}{A^2-169}}\\ \small A^2&=\small 181.25\\ \small A&=\small 13.5\,cm\\ \\ \small A\,in\,(1),\\ \\ \small 25&=\small \omega\sqrt{181.25-25}\\ \small \omega&=\small 2\,rads^{-1}\\ \small \omega =2&=\small \frac{2\pi}{T}\\ \small T&=\small \pi=3.1\,s \end{aligned}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS