Question

A body of mass 64 g is made to oscillate turn by turn on two different springs A and B. Spring A and B has force constant 4 `"N"/"m"` and `16 "N"/"m"` respectively. If T₁ and T₂ are period of m oscillations of springs A and B respectively, then `("T"_1 + "T"_2)/("T"_1 - "T"_2)` will be ______.

विकल्प

• 1 : 2

• 1 : 3

• 3 : 1

• 2 : 1

Answer 1

A body of mass 64 g is made to oscillate turn by turn on two different springs A and B. Spring A and B has force constant 4 `"N"/"m"` and `16 "N"/"m"` respectively. If T₁ and T₂ are period of m oscillations of springs A and B respectively, then `("T"_1 + "T"_2)/("T"_1 - "T"_2)` will be 3 : 1.

Explanation:

Given, m = 64 g = 64 x 10^-3 kg,

k_A = 4 N/m and k_B = 16 N/m

The time period of oscillation of a spring,

T = `2pi sqrt("m"/"k")`

`=> "T"_"A" = "T"_1 = 2pi sqrt("m"/"k"_"A")`

`= 2pi sqrt((64 xx 10^-3)/4)`

`= 2pi sqrt(16 xx 10^-3)`    ...(i)

and `"T"_"B" = "T"_2 = 2pi sqrt("m"/"k"_"B")`

`= 2pi sqrt((64 xx 10^-3)/16)`

`= 2pi sqrt(4 xx 10^-3)`    ...(ii)

Dividing Eq. (i) by Eq. (ii), we get

`"T"_1/"T"_2 = sqrt((16 xx 10^-3)/(4 xx 10^-3))` = 2

⇒ T₁ = 2T₂

`therefore ("T"_1 + "T"_2)/("T"_1 - "T"_2) = (3 "T"_2 + "T"_2)/(3 "T"_2 - "T"_2)`

`= "3T"_2/"T"_2 = 3/1` or 3 : 1