JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 22)
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 _________________.
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 _________________.
Answer
2
Explanation
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
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
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