Question 113: Oscillations

Question

A particle executes simple harmonic motion and is located at x = a, b and c at times t
0
, 2t
0
and 3t
0
respectively.  The freqquency of the oscillation is :

Accepted Answer

In general equation of simple harmonic motion, y = A sin $$\omega $$t
$$	herefore\,\,\,$$ a = A sin $$\omega $$t
0
$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t
0
$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t
0
a + c = A[sin $$\omega $$t
0
+ sin 3$$\omega $$t
0
]
= 2A sin 2$$\omega $$t
0
cos$$\omega $$t
0
$$ \Rightarrow $$$$\,\,\,$$a + c = 2 b cos$$\omega $$t
0
$$ \Rightarrow $$$$\,\,\,$$ $${{a + c} \over b}$$ = 2 cos$$\omega $$t
0
$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = $${1 \over {{t_0}}}$$ cos
$$-$$1
$$\left( {{{a + c} \over {2b}}} ight)$$
$$	herefore\,\,\,$$ f = $${\omega  \over {2\pi }}$$
= $${1 \over {2\pi {t_0}}}$$ cos
$$-$$1
$$\left( {{{a + c} \over {2b}}} ight)$$

Answer

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)$$

Answer

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)$$

Answer

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)$$

Answer

$${1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)$$

Oscillations

A particle executes simple harmonic motion and is located at x = a, b and c at times t 0 , 2t 0 and 3t 0 respectively. The freqquency of the oscillation is :

Choose the correct answer: