JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 14)
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1 . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is $${\pi \over 4}$$.
500 m/s2
500 $$\sqrt 2 m/$$ s2
750 m/s2
750 $$\sqrt 2 $$m / s2
Explanation
Mximum velocity, Vmax = a$$\omega $$
Maximum acceleration, Amax = a$$\omega $$2
Given that,
$${{a{\omega ^2}} \over {a\omega }}$$ = 10
$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = 10 s$$-$$1
Displacement, x = a sin ($$\omega $$t + $${\pi \over 4}$$)
at t = 0, displacement x = 5
$$\therefore\,\,\,$$ 5 = a sin $$\left( {{\pi \over 4}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ 5 = a $$ \times $$ $${1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$$$\,\,\,$$ a = 5$${\sqrt 2 }$$
$$\therefore\,\,\,$$ Maximum acceleration,
Amax = a$$\omega $$2 = 5 $${\sqrt 2 }$$ $$ \times $$ (10)2
= 500 $${\sqrt 2 }$$ m/s2
Maximum acceleration, Amax = a$$\omega $$2
Given that,
$${{a{\omega ^2}} \over {a\omega }}$$ = 10
$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = 10 s$$-$$1
Displacement, x = a sin ($$\omega $$t + $${\pi \over 4}$$)
at t = 0, displacement x = 5
$$\therefore\,\,\,$$ 5 = a sin $$\left( {{\pi \over 4}} \right)$$
$$ \Rightarrow $$$$\,\,\,$$ 5 = a $$ \times $$ $${1 \over {\sqrt 2 }}$$
$$ \Rightarrow $$$$\,\,\,$$ a = 5$${\sqrt 2 }$$
$$\therefore\,\,\,$$ Maximum acceleration,
Amax = a$$\omega $$2 = 5 $${\sqrt 2 }$$ $$ \times $$ (10)2
= 500 $${\sqrt 2 }$$ m/s2
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