Physics - SHM Question with Solution | TestHub

The speed of a particle executing simple harmonic motion is $v=\omega \sqrt{a^2-x^2}$ where, $a=$ Amplitude $\omega =$ Angular frequency $x=$ Displacement or $$v^2={\omega }^2\left(a^2-x^2\right)$$ According to the question, Here $$\begin{array}{rl}{v}_1^2& ={\omega }^2\left(a^2-x_1^2\right)\\ v_2^2& ={\omega }^2\left(a^2-x_2^2\right)\\ v_1^2-v_2^2& ={\omega }^2\left(x_2^2-x_1^2\right)\\ \omega & =\sqrt{\frac{v_1^2-v_2^2}{x_2^2-x_1^2}}\\ v_1=& 13\mathrm{m}/\mathrm{s}\end{array}$$ $$\begin{array}{rl}{v}_2& =12\mathrm{m}/\mathrm{s}\\ x_1& =3\mathrm{m}\\ x_2& =5\mathrm{m}\\ \omega & =\sqrt{\frac{(13{)}^2-(12{)}^2}{(5{)}^2-(3{)}^2}}\\ & =\sqrt{\frac{169-144}{25-9}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\\ \omega & =2\pi r\\ v& =\frac{\omega }{2\pi }=\frac{5/4}{2\pi }\\ v& =\frac{5}{8\pi }\end{array}$$