NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A particle executes simple harmonic motion and it is located at x = a , b and c at time t 0 , 2 t 0 a n d 3 t 0 respectively. The frequency of the oscillation is:
Options
- A1 2 π t 0 cos - 1 a + c 2 b
- B1 2 π t 0 cos - 1 a + 2 b 3 c
- C1 2 π t 0 cos - 1 a + b 2 c
- D1 2 π t 0 cos - 1 2 a + 3 c b
Correct answer
A. 1 2 π t 0 cos - 1 a + c 2 b
Step-by-step solution
a = A sin ⁡ ω t 0 ... ( 1 ) b = A sin ⁡ 2 ω t 0 ... ( 2 ) c = A sin ⁡ 3 ω t 0 ... ( 3 ) adding equation ( 1 ) and ( 3 ) a + c = A sin ⁡ ω t 0 + sin ⁡ 3 ω t 0 = 2 A sin ⁡ 2 ω t 0 cos ⁡ ω t 0 ... ( 4 ) ∵     sin C + sin D =    2 sin C + D 2 cos C − D 2 from equation ( 2 ) and ( 4 ) a + c b = 2 cos ω t 0 ω =   1 t 0 cos - 1 ⁡ a + c 2 b   ⇒ f = 1 2 π t 0 cos - 1 ⁡ a