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Question: A particle executes simple harmonic motion and is located at \( x = a,b \) and \( c \) at times \( {...

A particle executes simple harmonic motion and is located at x=a,bx = a,b and cc at times t0,2t0{t_{0,}}2{t_0} and 3t03{t_0} respectively. The frequency of the oscillation is
(A) 12πt0cos1(2a+3cb)\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{2a + 3c}}{b}} \right)
(B) 12πt0cos1(a+c2b)\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right)
(C) 12πt0cos1(a+c2c)\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2c}}} \right)
(D) 12πt0cos1(a+2b2c)\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + 2b}}{{2c}}} \right)

Explanation

Solution

Hint : to solve this problem we should know about simple harmonic motion. The general SHM equation applies to all simple oscillating motion is,
x=x0cos(ωt)x = {x_0}\cos (\omega t)
Here, x0{x_0} is the amplitude of the SHM and ω\omega is the angular frequency of the SHM.

Complete Step By Step Answer:
A particle executes simple harmonic motion having AA is amplitude of SHM and ω\omega is angular frequency of the SHM.
As per location given in question. Equation of simple harmonic motion will be at x=a,bx = a,b and cc at given time t0,2t0{t_{0,}}2{t_0} and 3t03{t_0} will be respectively,
a=Acosωt0a = A\cos \omega {t_0} ………………………… (1)(1)
b=Acos2ωt0b = A\cos 2\omega {t_0} ………………………… (2)(2)
c=Acos3ωt0c = A\cos 3\omega {t_0} ………………………… (3)(3)
On adding (1)(1) and (3)(3) . We get,
a+c=A(cosωt0+cos3ωt0)a + c = A(\cos \omega {t_0} + \cos 3\omega {t_0})
By applying a trigonometric equation.
a+c=2A(cos(3ωt0+ωt02)cos(3ωt0ωt02))\Rightarrow a + c = 2A\left( {\cos \left( {\dfrac{{3\omega {t_0} + \omega {t_0}}}{2}} \right)\cos \left( {\dfrac{{3\omega {t_0} - \omega {t_0}}}{2}} \right)} \right)
a+c=2Acos2ωt0cosωt0\Rightarrow a + c = 2A\cos 2\omega {t_0}\cos \omega {t_0}
From (2)(2) we get,
b=Acos2ωt0b = A\cos 2\omega {t_0}
a+c=2bcosωt0\Rightarrow a + c = 2b\cos \omega {t_0}
By taking the inverse. We get,
cos1(a+c2b)=ωt0\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) = \omega {t_0}
As we know ω=2πf\omega = 2\pi f . So,
cos1(a+c2b)=2πft0\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) = 2\pi f{t_0}
f=12πt0cos1(a+c2b)\Rightarrow f = \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right)
Hence, (b) 12πt0cos1(a+c2b)\dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) is correct option.

Note :
In simple harmonic motion particles oscillate about their mean position about which particle is to its to and for motion. In simple harmonic motion distance is directly proportional to acceleration of the particle. The maximum kinetic energy of a particle is at mean position and maximum potential energy is at maximum position and vice-versa.