Lock down period

Classical Mechanics Level 3

A particle executes S. H. M. starting from one of it's extreme positions. In three successive seconds, it's distances from the mean position are 4 , 7 , 8 4,7,8 units. What is the time period of motion of the particle?


The answer is 11.612.

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1 solution

Let the amplitude of motion be A A and the time period be T T . Then the equation of motion of the particle is

x = A cos ⁡ ( 2 π t T ) x=A\cos (\frac{2πt}{T}) .

Let the three successive times be t − 1 , t , t + 1 t-1,t,t+1 secs. Then

4 = A cos ⁡ ( 2 π ( t − 1 ) T ) 4=A\cos \left (\dfrac{2π(t-1)}{T}\right )

7 = A cos ⁡ ( 2 π t T ) 7=A\cos (\frac{2πt}{T})

8 = A cos ⁡ ( 2 π ( t + 1 ) T ) 8=A\cos \left (\dfrac{2π(t+1)}{T}\right ) .

So 4 + 8 = 12 = 2 A cos ⁡ ( 2 π t T ) cos ⁡ ( 2 π T ) = 14 cos ⁡ ( 2 π T ) 4+8=12=2A\cos (\frac{2πt}{T})\cos (\frac{2π}{T})=14\cos (\frac{2π}{T})

⟹ T = 2 π cos ⁡ − 1 ( 6 7 ) ≈ 11.612 \implies T=\dfrac{2π}{\cos^{-1} (\frac{6}{7})}\approx \boxed {11.612} .

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