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 = 2p $\sqrt { \frac { \mathrm { m } } { \mathrm { k } _ { 1 } } }$ ….(i), t₂= 2p $\sqrt { \frac { \mathrm { m } } { \mathrm { k } _ { 2 } } }$ …..(ii)

when springs are in series then

T = 2p = 2p $\sqrt { \frac { m \left( k _ { 1 } + k _ { 2 } \right) } { k _ { 1 } k _ { 2 } } }$

Squaring and adding (1) and (2) we get

t₁² + t₂² = 4p² $\frac { \mathrm { m } } { \mathrm { k } _ { 1 } } + 4 \pi ^ { 2 } \frac { \mathrm {~m} } { \mathrm { k } _ { 2 } }$ = 4p²m

or t₁² + t­₂² = T²