The time period of a simple pendulum is T1 when its point of suspension is at rest with respect to the ground. When the point of suspension moves vertically upwards according to the law y=kt2 where k=1 ms- 2 , its new time period is T2 , then the value of (T12/T22) is [. Take g=10 ms- 2].

Question

The time period of a simple pendulum is T1 when its point of suspension is at rest with respect to the ground. When the point of suspension moves vertically upwards according to the law y=kt2 where k=1 ms- 2 , its new time period is T2 , then the value of (T12/T22) is [. Take g=10 ms- 2]. (A) (4/5) (B) (6/5) (C) (5/6) (D) 1

Tardigrade Admin · Accepted Answer

Acceleration of the point of suspension a=(d2 y/d t2)=2k=2m / s2 T=2π √(L/ge f f) ⇒ T1=2π √(L/ textg) T2=2π √(L/g + 2) (T12/T22)=(g + 2/g)=(10 + 2/10)=(6/5)

$\frac{4}{5}$

$\frac{6}{5}$

$\frac{5}{6}$

$1$

Acceleration of the point of suspension
$a = \frac{d ^{2} y}{d t ^{2}} = 2 k = 2 m / s^{2}$
$T = 2 π \frac{L}{g _{e ff}} \Rightarrow T_{1} = 2 π \frac{L}{g}$
$T_{2} = 2 π \frac{L}{g + 2}$
$\frac{T _{1}^{2}}{T _{2}^{2}} = \frac{g + 2}{g} = \frac{10 + 2}{10} = \frac{6}{5}$

54​

56​

65​

1

Acceleration of the point of suspension a=dt2d2y​=2k=2m/s2 T=2πgeff​L​​⇒T1​=2πgL​​ T2​=2πg+2L​​ T22​T12​​=gg+2​=1010+2​=56​

$\frac{4}{5}$

$\frac{6}{5}$

$\frac{5}{6}$

$1$

Acceleration of the point of suspension
$a = \frac{d ^{2} y}{d t ^{2}} = 2 k = 2 m / s^{2}$
$T = 2 π \frac{L}{g _{e ff}} \Rightarrow T_{1} = 2 π \frac{L}{g}$
$T_{2} = 2 π \frac{L}{g + 2}$
$\frac{T _{1}^{2}}{T _{2}^{2}} = \frac{g + 2}{g} = \frac{10 + 2}{10} = \frac{6}{5}$