A man measures the period of a simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates upwards with an acceleration g / 4, then the period of pendulum will be

Question

A man measures the period of a simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates upwards with an acceleration g / 4, then the period of pendulum will be (A) 2 T √5 (B) T (C) (2 T/√5) (D) (T/4)

Tardigrade Admin · Accepted Answer

As, T=2 π √(l/g). When lift is accelerated upwards with acceleration a (=g / 4), then effective acceleration due to gravity inside the lift g1=g+a=g+(g/4)=(5 g/4) ∴ T1=2 π √(l/5 g / 4)=2 π √(l/g) × (2/√5)=(2 T/√5)

Q. A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ second. If the lift accelerates upwards with an acceleration $g /4$ , then the period of pendulum will be

$2 T 5$

$T$

$\frac{2 T}{5}$

$\frac{T}{4}$

As, $T = 2 π \frac{l}{g}$ . When lift is accelerated upwards with acceleration a $(= g /4)$ , then effective acceleration due to gravity inside the lift
$g_{1} = g + a = g + \frac{g}{4} = \frac{5 g}{4}$
$∴ T_{1} = 2 π \frac{l}{5 g /4} = 2 π \frac{l}{g} \times \frac{2}{5} = \frac{2 T}{5}$

Q. A man measures the period of a simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates upwards with an acceleration g/4, then the period of pendulum will be

2T5​

T

5​2T​

4T​

As, T=2πgl​​. When lift is accelerated upwards with acceleration a (=g/4), then effective acceleration due to gravity inside the lift g1​=g+a=g+4g​=45g​ ∴T1​=2π5g/4l​​=2πgl​​×5​2​=5​2T​

Q. A man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ second. If the lift accelerates upwards with an acceleration $g /4$ , then the period of pendulum will be

$2 T 5$

$T$

$\frac{2 T}{5}$

$\frac{T}{4}$

As, $T = 2 π \frac{l}{g}$ . When lift is accelerated upwards with acceleration a $(= g /4)$ , then effective acceleration due to gravity inside the lift
$g_{1} = g + a = g + \frac{g}{4} = \frac{5 g}{4}$
$∴ T_{1} = 2 π \frac{l}{5 g /4} = 2 π \frac{l}{g} \times \frac{2}{5} = \frac{2 T}{5}$