Question

The bob of a simple pendulum performs SHM with period T in air and with period T₁ in water Relation between T and T₁ is ______.

(neglect friction due to water, density of the material of the bob is = `9/8xx10^3` kgm³, density of water = 1 gcc^-1) 

पर्याय

• T₁ = 3T

• T₁ = 2T

• T₁ = T

• T₁ = `"T"/2`

Answer 1

The bob of a simple pendulum performs SHM with period T in air and with period T₁ in water Relation between T and T₁ is T₁ = 3T.

Explanation:

Time period of simple pendulum in water (T) is given by

T = 2π `sqrt(1/"g"_"eff")`

[g_eff = acceleration due to gravity in water]

When a bob is submerged in water, the effective value of gravity's acceleration is given by

As we know g_eff = g `((sigma-rho)/sigma)`

[Here, σ = density of bob, ρ = density of water]

= 9.8`((9/8xx10^3-10^3)/(9/8xx10^3))`

= 9.8`((9/8-1)/(9/8))`

= 9.8`(1/8xx8/9)`

= `9.8/9`

So, T₁ = 2π `sqrt(("l"xx9)/9.8)`

⇒ T₁ = 3T                      `[therefore "T" = 2pi sqrt(ℓ/g)]`