Question 146: Oscillations

Question

A particle performs simple harmonic motion with amplitude $$A.$$ Its speed is trebled at the instant that it is at a distance $${{2A} \over 3}$$ from equilibrium position. The new amplitude of the motion is:

Accepted Answer

We know that $$V = \omega \sqrt {{A^2} - {x^2}} $$
Initially $$v = \omega \sqrt {{A^2} - {{\left( {{{2A} \over 3}} ight)}^2}} $$
Finally $$3v = \omega \sqrt {A_{new}^2 - {{\left( {{{2A} \over 3}} ight)}^2}} $$
where $${A_{new}}$$ = final amplitude (Given at $$x = {{2A} \over 3},$$ velocity to trebled)
On dividing we get $${3 \over 1} = {{\sqrt {A_{new}^2 - {{\left( {{{2A} \over 3}} ight)}^2}} } \over {\sqrt {{A^2} - {{\left( {{{2A} \over 3}} ight)}^2}} }}$$
$$9\left[ {{A^2} - {{4{A^2}} \over 9}} ight] = A_{new}^2 - {{4{A^2}} \over 9}$$
$$	herefore$$ $$A_{new} = {{7A} \over 3}$$

Answer

$$A\sqrt 3 $$

Answer

$${{7A} \over 3}$$

Answer

$${A \over 3}\sqrt {41} $$

Answer

$$3A$$

Oscillations

A particle performs simple harmonic motion with amplitude $A. Its speed is trebled at the instant that it is at a distance {{2A} \over 3}$ from equilibrium position. The new amplitude of the motion is:

Choose the correct answer: