Verified Solution • 2025–26 Session moderate

QJEEM18Phb522: JEE Main 2018 Physics - Class 11 JEEM18Phb522 Solution: A particle executes simple harmonic motion and is located at x…

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 :

Multiple Choice Options

  • A 12 π t0cos⁡−1(a+c2b){1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + c} \over {2b}}} \right)
  • B 12 π t0cos⁡−1(a+b2c){1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + b} \over {2c}}} \right)
  • C 12 π t0cos⁡−1(2a+3cb){1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{2a + 3c} \over b}} \right)
  • D 12 π t0cos⁡−1(a+2b3c){1 \over {2\,\pi \,{t_0}}}{\cos ^{ - 1}}\left( {{{a + 2b} \over {3c}}} \right)

Step-by-Step Verified Solution

In general equation of simple harmonic motion, y = A sin ω\omega t

∴   \therefore\,\,\, a = A sin ω\omega t0

      \,\,\,\,\,\, b = A sin 2 ω\omega t0

       \,\,\,\,\,\,\, c = A sin 3 ω\omega t0

a + c = A[sin ω\omega t0 + sin 3 ω\omega t0]

= 2A sin 2 ω\omega t0 cos ω\omega t0

⇒\Rightarrow    \,\,\, a + c = 2 b cos ω\omega t0

⇒\Rightarrow    \,\,\, a+cb{{a + c} \over b} = 2 cos ω\omega t0

⇒\Rightarrow    \,\,\, ω\omega = 1t0{1 \over {{t_0}}} cos −- 1 (a+c2b)\left( {{{a + c} \over {2b}}} \right)

∴   \therefore\,\,\, f = ω2π{\omega \over {2\pi }}

= 12πt0{1 \over {2\pi {t_0}}} cos −- 1 (a+c2b)\left( {{{a + c} \over {2b}}} \right)

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