Answer to Question #184082 in Physics for sherry

Question

For a particle executing SHM the displacement is 8 cm at that instant the velocity is 6 cm/s and the

displacement is 6 cm at that instant the velocity is 8 cm/s. Calculate (i) amplitude(A), (ii) frequency(f)

and (iii) time period(T).

Answer 1

i)

$$v_1=\omega\sqrt{A^2-x_1^2}\\v_2=\omega\sqrt{A^2-x_2^2}\\\frac{v_1}{v_2}=\frac{\sqrt{A^2-x_1^2}}{\sqrt{A^2-x_2^2}}\\ \frac{6}{8}=\frac{\sqrt{A^2-8^2}}{\sqrt{A^2-6^2}}\\A=10\ cm$$

ii)

$$6=2\pi f\sqrt{10^2-8^2}\\f=0.16\ Hz$$

iii)

$$\\T=\frac{1}{0.16}=6.3\ s$$