Question

A particle at the end of a spring executes simple harmonic motion with a period t₁, while the corresponding period for another spring is t₂. If the period of oscillation with the two springs in series is T, then -

• A. T = t₁ + t₂
• B. T² = t₁² + t₂²
• C. T^–1 = t₁^–1 + t₂^–1
• D. T^–2 = t₁^–2 + t₂^–2

Answer 1

Answer: (B)

T² = t₁² + t₂²

Answer 2

t₁ = 2π$\sqrt{\frac{m}{k_{1}}}$ ….(i), t₂ = 2π$\sqrt{\frac{m}{k_{2}}}$….(ii)

when springs are in series then

T = 2π $\sqrt{\frac{m}{\frac{k_{1}k_{2}}{k_{1} + k_{2}}}}$ = 2π$\sqrt{\frac{m(k_{1} + k_{2})}{k_{1}k_{2}}}$

squaring and adding (i) and (ii) we get

t₁² + t₂² = 4π²$\frac{m}{k_{1}}$ + 4π²$\frac{m}{k_{2}}$

= 4π²m $\left( \frac{k_{1} + k_{2}}{k_{1}k_{2}} \right)$ or

t₁² + t₂² = T²