"A particle executes simple harmonic motion represented by displacement function as x(t) = A sin($\omega$t + $\phi$)If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s$-$1 respectively, then its amplitude is $x\sqrt 2$ cm where the value of x is _________________."

Question

Accepted Answer

"x(t) = A sin($\omega$t + $\phi$)

v(t) = A$\omega$ cos ($\omega$t + $\phi$)

2 = A sin$\phi$ ...... (1)

2$\omega$ = A$\omega$ cos$\phi$ ....... (2)

From (1) and (2)

tan$\phi$ = 1

$\phi$ = 45$^\circ$

Putting value of $\phi$ in equation (1),

$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$

$A = 2\sqrt 2$

$\therefore$ x = 2"

A particle executes simple harmonic motion represented by displacement function as

x(t) = A sin($\omega$t + $\phi$)

If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s^$-$1 respectively, then its amplitude is $x\sqrt 2$ cm where the value of x is _________________.

Solution

About this question

Drill 25 more like these. Every day. Free.

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin($\omega$t + $\phi$)If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s$-$1 respectively, then its amplitude is $x\sqrt 2$ cm where the value of x is _________________.

Solution

About this question

More Oscillations questions

Drill 25 more like these. Every day. Free.

A particle executes simple harmonic motion represented by displacement function as

x(t) = A sin($\omega$t + $\phi$)

If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s^$-$1 respectively, then its amplitude is $x\sqrt 2$ cm where the value of x is _________________.