A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance (3 l/4) . The pendulum is released from rest. Throughout the motion the moving string makes small angles with vertical. Time after which the pendulum returns back to its initial position is -

Question

A pendulum has time period T for small oscillations. An obstacle P is situated below the point of suspension O at a distance (3 l/4) . The pendulum is released from rest. Throughout the motion the moving string makes small angles with vertical. Time after which the pendulum returns back to its initial position is - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jfvrviosgiojkwyu.jpg" /> (A) T (B) 3 T / 4 (C) 3 T / 5 (D) 4 T / 5

Tardigrade Admin · Accepted Answer

first half oscillation is covered with length

l

and second half with length

\frac{3 l}{4}

therefore total time can be measured by

t = \frac{T _{1}}{2} + \frac{T _{2}}{2}

T_{1} = 2 π \frac{1}{g} = T

T_{2} = 2 π \frac{1/4}{g} = \frac{T}{2}

Putting value we get

t = \frac{3 T}{4}

$T$

$3 T /4$

$3 T /5$

$4 T /5$

first half oscillation is covered with length $l$ and second half with length $\frac{3 l}{4}$ therefore total time can be measured by
$t = \frac{T _{1}}{2} + \frac{T _{2}}{2}$
$T_{1} = 2 π \frac{1}{g} = T$
$T_{2} = 2 π \frac{1/4}{g} = \frac{T}{2}$
Putting value we get $t = \frac{3 T}{4}$

T

3T/4

3T/5

4T/5

first half oscillation is covered with length l and second half with length 43l​ therefore total time can be measured by t=2T1​​+2T2​​​ T1​=2πg1​​=T T2​=2πg1/4​​=2T​ Putting value we get t=43T​

$T$

$3 T /4$

$3 T /5$

$4 T /5$

first half oscillation is covered with length $l$ and second half with length $\frac{3 l}{4}$ therefore total time can be measured by
$t = \frac{T _{1}}{2} + \frac{T _{2}}{2}$
$T_{1} = 2 π \frac{1}{g} = T$
$T_{2} = 2 π \frac{1/4}{g} = \frac{T}{2}$
Putting value we get $t = \frac{3 T}{4}$