The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44 %. then its new time period of oscillation will be

Question

The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44 %. then its new time period of oscillation will be (A) 96 s (B) 58 s (C) 82 s (D) 72 s

Tardigrade Admin · Accepted Answer

Let initial length be l1 Final length l2=l1 × (144/100) T1=2 π √(l1/g) T2=2 π √(l1/g) × (144/100) or T2 = 1.2 T1 T1=60 s So T2=72 s

Q. The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44%$ . then its new time period of oscillation will be

96 s

58 s

82 s

72 s

Let initial length be $l_{1}$
Final length $l_{2} = l_{1} \times \frac{144}{100}$
$T_{1} = 2 π \frac{l _{1}}{g}$
$T_{2} = 2 π \frac{l _{1}}{g} \times \frac{144}{100}$
or $T_{2} = 1.2 T_{1}$
$T_{1} = 60 s$
So $T_{2} = 72 s$

Q. The time period of oscillations of a simple pendulum is 1 minute. If its length is increased by 44%. then its new time period of oscillation will be