A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = kt2, (k = 1ms-2) where y is the vertical displacement. The time period now becomes T2. The ratio of (T21/T22) is (Take g = 10ms-2)

Question

A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = kt2, (k = 1ms-2) where y is the vertical displacement. The time period now becomes T2. The ratio of (T21/T22) is (Take g = 10ms-2) (A) 6/5 (B) 5/6 (C) 1 (D) 4/5

Tardigrade Admin · Accepted Answer

y=kt2 ⇒ (d2y/dt2)=2k ay=2ms/s2 (as k=1m/s2) T1=2π√(l/g) and T2=2π√(l/g+ay) ∴ (T21/T22)=(g+ay/g)=(10+2/10)=(6/5) ∴ Correct answer is (a)

Q. A simple pendulum has time period T1​. The point of suspension is now moved upward according to the relation y=kt2,(k=1ms−2) where y is the vertical displacement. The time period now becomes T2​. The ratio of T22​T12​​ is (Takeg=10ms−2)

6/5

5/6

1

4/5

y=kt2⇒dt2d2y​=2k ay​=2ms/s2 (ask=1m/s2) T1​=2πgl​​ and T2​=2πg+ay​l​​ ∴T22​T12​​=gg+ay​​=1010+2​=56​ ∴ Correct answer is (a)

$y = k t^{2} \Rightarrow \frac{d ^{2} y}{d t ^{2}} = 2 k$
$a_{y} = 2 m s / s^{2}$ $(a s k = 1 m / s^{2})$
$T_{1} = 2 π \frac{l}{g}$
and $T_{2} = 2 π \frac{l}{g + a _{y}}$
$∴ \frac{T _{1}^{2}}{T _{2}^{2}} = \frac{g + a _{y}}{g} = \frac{10 + 2}{10} = \frac{6}{5}$
$∴$ Correct answer is (a)