JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 13)
A particle executes simple harmonic motion and is located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The freqquency of the oscillation is :
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$
$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$
Explanation
In general equation of simple harmonic motion, y = A sin $$\omega $$t
$$\therefore\,\,\,$$ a = A sin $$\omega $$t0
$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t0
$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t0
a + c = A[sin $$\omega $$t0 + sin 3$$\omega $$t0]
= 2A sin 2$$\omega $$t0 cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$
$$\therefore\,\,\,$$ f = $${\omega \over {2\pi }}$$
= $${1 \over {2\pi {t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$
$$\therefore\,\,\,$$ a = A sin $$\omega $$t0
$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t0
$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t0
a + c = A[sin $$\omega $$t0 + sin 3$$\omega $$t0]
= 2A sin 2$$\omega $$t0 cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t0
$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$
$$\therefore\,\,\,$$ f = $${\omega \over {2\pi }}$$
= $${1 \over {2\pi {t_0}}}$$ cos$$-$$1 $$\left( {{{a + c} \over {2b}}} \right)$$
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