Question

A particle executes SHM between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T₁ and to go from A/2 to A is T₂. Then

• A. T₁< T₂
• B. T₁> T₂
• C. T₁ = T₂
• D. T₁ = 2T₂

Answer 1

Answer: (C)

T₁ = T₂

Answer 2

Let x = A sin ωt

From 0 – A/2

∴ A/2 = A sin ωT₁

or sin ωT₁ = $\frac { 1 } { 2 }$ = sin $\frac { \pi } { 6 }$

∴ $\frac { 2 \pi } { T } \cdot T _ { 1 } = \frac { \pi } { 6 }$ or T₁ = $\frac { \mathrm { T } } { 12 }$

From 0 – A

A = A sin ωT’₁

or, sin ωT’₁ =1 = sin (π / 2)

or, $\frac { 2 \pi } { \mathrm { T } } \cdot \mathrm { T } _ { 1 } ^ { \prime } = \frac { \pi } { 2 }$ or T’₁ = $\frac { T } { 4 }$

∴ From A/2 – A :

T₂ = T’₁ − T₁ = = $\frac { T } { 6 }$

T₁ = and T₂ =

∴ T₁ < T₂