A particle executes S.H.M. between x = − A and x = + A . The time taken for it to go from mean position O to 2 A is T 1 and to go from 2 A to A is T 2 , then ( sin 3 0 ∘ = 1/2 , sin 9 0 ∘ = 1

Question

A particle executes S.H.M. between x = − A and x = + A . The time taken for it to go from mean position O to 2 A ​ is T 1 ​ and to go from 2 A ​ to A is T 2 ​ , then ( sin 3 0 ∘ = 1/2 , sin 9 0 ∘ = 1 )

TestHub · Accepted Answer

The equation of simple harmonic motion starting from the mean position is given by $x = A sin (ω t)$ .

For the particle to travel from the mean position $x = 0$ to $x = \frac{A}{2}$ , the time taken is $T_{1}$ .

$\frac{A}{2} = A sin (ω T_{1})$

$sin (ω T_{1}) = \frac{1}{2} \Rightarrow ω T_{1} = \frac{π}{6}$

$T_{1} = \frac{π}{6 ω}$

For the particle to travel from the mean position to $x = A$ , the total time taken is $T_{1} + T_{2}$ .

$A = A sin (ω (T_{1} + T_{2}))$

$sin (ω (T_{1} + T_{2})) = 1 \Rightarrow ω (T_{1} + T_{2}) = \frac{π}{2}$

$T_{1} + T_{2} = \frac{π}{2 ω}$

Substituting the value of $T_{1}$ :

$T_{2} = \frac{π}{2 ω} - \frac{π}{6 ω} = \frac{3 π - π}{6 ω} = \frac{2 π}{6 ω} = \frac{π}{3 ω}$

Comparing $T_{1}$ and $T_{2}$ , we get $T_{2} = 2 T_{1}$ .

Therefore, $T_{1} < T_{2}$ .

Answer: $T_{1} < T_{2}$

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