The bob of a simple pendulum executes SHM in water with a period T, while the period of oscillation of the bob is T0 in air. Neglecting frictional force of water and given that the density of the bob is ((4/3)) × 1000 kg m -3. The relationship between T and T0 is

Question

The bob of a simple pendulum executes SHM in water with a period T, while the period of oscillation of the bob is T0 in air. Neglecting frictional force of water and given that the density of the bob is ((4/3)) × 1000 kg m -3. The relationship between T and T0 is (A) T=T0 (B) T=4 T0 (C) T=2 T0 (D) T=(T0/2)

Tardigrade Admin · Accepted Answer

Here, density of bob, ρ=(4/3) × 1000 kg m -3 and density of water, σ=1000 kg m -3 In air, T0=2 π √(l/g) dots(i) In water, T=2 π √(l/g(1-(σ)ρ))=2 π √(l/g(1-(3)4)) =2(2 π √(l/g))=2 T0 (Using(i))

Q. The bob of a simple pendulum executes SHM in water with a period $T$ , while the period of oscillation of the bob is $T_{0}$ in air. Neglecting frictional force of water and given that the density of the bob is $(\frac{4}{3}) \times 1000 k g m^{- 3}$ . The relationship between $T$ and $T_{0}$ is

$T = T_{0}$

$T = 4 T_{0}$

$T = 2 T_{0}$

$T = \frac{T _{0}}{2}$

Here, density of bob, $ρ = \frac{4}{3} \times 1000 k g m^{- 3}$ and
density of water, $σ = 1000 k g m^{- 3}$
In air, $T_{0} = 2 π \frac{l}{g} \dots (i)$
In water, $T = 2 π \frac{l}{g ( 1 - \frac{σ}{ρ} )} = 2 π \frac{l}{g ( 1 - \frac{3}{4} )}$
$= 2 (2 π \frac{l}{g}) = 2 T_{0}$ (Using(i))

Q. The bob of a simple pendulum executes SHM in water with a period T, while the period of oscillation of the bob is T0​ in air. Neglecting frictional force of water and given that the density of the bob is (34​)×1000kgm−3. The relationship between T and T0​ is

T=T0​

T=4T0​

T=2T0​

T=2T0​​