The time period of a simple pendulum of length √5 m suspended in a car moving with uniform acceleration of a 5 ms-2 in a horizontal straight road is (g = 10 ms-2)

Question

The time period of a simple pendulum of length √5 m suspended in a car moving with uniform acceleration of a 5 ms-2 in a horizontal straight road is (g = 10 ms-2) (A) (2π/√5) s (B) (π/√5) s (C) 5 π s (D) 4 π s

Tardigrade Admin · Accepted Answer

Given, Length of simple pendulum, l=√5 m Acceleration of car (a)=5 ms -2 Now g text effective =√a2+g2 =√(5)2+(10)2 =√125=5 √5 T =2 π √(l/g text effective ) T =2 π √(√5/5 √5) T =2 π / √5 s

Q. The time period of a simple pendulum of length $5 m$ suspended in a car moving with uniform acceleration of a $5 m s^{- 2}$ in a horizontal straight road is $(g = 10 m s^{- 2})$

$\frac{2 π}{5} s$

$\frac{π}{5} s$

$5 π s$

$4 π s$

$3 π s$

Given, Length of simple pendulum, $l = 5 m$
Acceleration of car $(a) = 5 m s^{- 2}$
Now $g_{effective} = a^{2} + g^{2}$
$= (5)^{2} + (10)^{2}$
$= 125 = 55$
$T = 2 π \frac{l}{g _{effective}}$
$T = 2 π \frac{5}{5 5}$
$T = 2 π / 5 s$

Q. The time period of a simple pendulum of length 5​m suspended in a car moving with uniform acceleration of a 5ms−2 in a horizontal straight road is (g=10ms−2)

5​2π​s

5​π​s

5πs

4πs

3πs

Given, Length of simple pendulum, l=5​m Acceleration of car (a)=5ms−2 Now geffective ​=a2+g2​ =(5)2+(10)2​ =125​=55​ T=2πgeffective ​l​​ T=2π55​5​​​ T=2π/5​s