A simple pendulum of length l1, has a time period of 4 s and another simple pendulum of length l2 has a time period 3 s. Then the time period of another pendulum of length (l1-l2) is

Question

A simple pendulum of length l1, has a time period of 4 s and another simple pendulum of length l2 has a time period 3 s. Then the time period of another pendulum of length (l1-l2) is (A) √3 s (B) 1 s (C) √(3/4) s (D) √7 s

Tardigrade Admin · Accepted Answer

4 s =2 π √(l1/g) or l1=(4 g/π2) 3 s =2 π √(l2/g) or l2=(9 g/4 π2) (l1-l2)=(g/π2)(4-(9/4))=(7 g/4 π2) T prime=2 π √((l1-l2)/g)=2 π √(7 g/4 π2 g)=√7 s

Q. A simple pendulum of length $l_{1}$ , has a time period of $4 s$ and another simple pendulum of length $l_{2}$ has a time period $3 s$ . Then the time period of another pendulum of length $(l_{1} - l_{2})$ is

$3 s$

$1 s$

$\frac{3}{4} s$

$7 s$

$4 s = 2 π \frac{l _{1}}{g}$ or $l_{1} = \frac{4 g}{π ^{2}}$
$3 s = 2 π \frac{l _{2}}{g}$ or $l_{2} = \frac{9 g}{4 π ^{2}}$
$(l_{1} - l_{2}) = \frac{g}{π ^{2}} (4 - \frac{9}{4}) = \frac{7 g}{4 π ^{2}}$
$T^{'} = 2 π \frac{( l _{1} - l _{2} )}{g} = 2 π \frac{7 g}{4 π ^{2} g} = 7 s$

Q. A simple pendulum of length l1​, has a time period of 4s and another simple pendulum of length l2​ has a time period 3s. Then the time period of another pendulum of length (l1​−l2​) is

3​s

1s

43​​s

7​s