A simple pendulum has time period 2 s. The point of suspension is now moved upward according to relation y=(6 t-3.75 t2) m where t is in second and y is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be (g=10 ms -2)

Question

A simple pendulum has time period 2 s. The point of suspension is now moved upward according to relation y=(6 t-3.75 t2) m where t is in second and y is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be (g=10 ms -2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T1=2 s =2 π √(l/g), T2=2 π √(l/g prime) where g'=g+(d2 y/d t2)=10-7.5=2.5=(g/4) ⇒ T2=2 π √(l/g / 4)=2 × 2=4 s

Q. A simple pendulum has time period $2 s$ . The point of suspension is now moved upward according to relation $y = (6 t - 3.75 t^{2}) m$ where $t$ is in second and $y$ is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be $(g = 10 m s^{- 2})$

$T_{1} = 2 s = 2 π \frac{l}{g}, T_{2} = 2 π \frac{l}{g ^{'}}$
where $g^{'} = g + \frac{d ^{2} y}{d t ^{2}} = 10 - 7.5 = 2.5 = \frac{g}{4}$
$\Rightarrow T_{2} = 2 π \frac{l}{g /4} = 2 \times 2 = 4 s$

Q. A simple pendulum has time period 2s. The point of suspension is now moved upward according to relation y=(6t−3.75t2)m where t is in second and y is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be (g=10ms−2)

T1​=2s=2πgl​​,T2​=2πg′l​​ where g′=g+dt2d2y​=10−7.5=2.5=4g​ ⇒T2​=2πg/4l​​=2×2=4s

Q. A simple pendulum has time period $2 s$ . The point of suspension is now moved upward according to relation $y = (6 t - 3.75 t^{2}) m$ where $t$ is in second and $y$ is the vertical displacement in upward direction. The new time period (in s) of simple pendulum will be $(g = 10 m s^{- 2})$

$T_{1} = 2 s = 2 π \frac{l}{g}, T_{2} = 2 π \frac{l}{g ^{'}}$
where $g^{'} = g + \frac{d ^{2} y}{d t ^{2}} = 10 - 7.5 = 2.5 = \frac{g}{4}$
$\Rightarrow T_{2} = 2 π \frac{l}{g /4} = 2 \times 2 = 4 s$