There is a rod of length l and mass m It is hinged at one end to the ceiling. The period of small oscillations is

Question

There is a rod of length l and mass m It is hinged at one end to the ceiling. The period of small oscillations is (A) T=2 π √((2 l/3 g)) (B) T=π √((l/3 g)) (C) T=2 π √((l/3 g)) (D) T=2 π √((l/g))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f8cacde63cc75265f16da7948ebd2c0f-.png" /> As T=2 π √((I/m g h)) Here, I=(m l2/3) and h=(l/2) ∴ T=2 π √[(((m l2/3))/m g (l)2)]=2 π √((2 l/3 g))

Q. There is a rod of length l and mass m It is hinged at one end to the ceiling. The period of small oscillations is

$T = 2 π (\frac{2 l}{3 g})$

$T = π (\frac{l}{3 g})$

$T = 2 π (\frac{l}{3 g})$

$T = 2 π (\frac{l}{g})$

As $T = 2 π (\frac{I}{m g h})$
Here, $I = \frac{m l ^{2}}{3}$ and $h = \frac{l}{2}$
$∴ T = 2 π [\frac{( \frac{m l ^{2}}{3} )}{m g \frac{l}{2}}] = 2 π (\frac{2 l}{3 g})$

Q. There is a rod of length l and mass m It is hinged at one end to the ceiling. The period of small oscillations is

T=2π(3g2l​)​

T=π(3gl​)​

T=2π(3gl​)​

T=2π(gl​)​

As T=2π(mghI​)​ Here, I=3ml2​ and h=2l​ ∴T=2π[mg2l​(3ml2​)​]​=2π(3g2l​)​

$T = 2 π (\frac{2 l}{3 g})$

$T = π (\frac{l}{3 g})$

$T = 2 π (\frac{l}{3 g})$

$T = 2 π (\frac{l}{g})$

As $T = 2 π (\frac{I}{m g h})$
Here, $I = \frac{m l ^{2}}{3}$ and $h = \frac{l}{2}$
$∴ T = 2 π [\frac{( \frac{m l ^{2}}{3} )}{m g \frac{l}{2}}] = 2 π (\frac{2 l}{3 g})$