Question

A particle executing SHM has velocities v₁ and v₂ at distances x₁ and x₂ respectively, from the mean position. Its time period is ______.

विकल्प

• `2pisqrt((x_2^2 - x_1^2)/("v"_2^2 - "v"_1^2))`

• `2pisqrt((x_1^2 - x_2^2)/("v"_2^2 - "v"_1^2))`

• `2pisqrt((x_1x_2)/("v"_2^2 - "v"_1^2))`

• `2pisqrt((x_1 + x_2)/("v"_1"v"_2))`

Answer 1

A particle executing SHM has velocities v₁ and v₂ at distances x₁ and x₂ respectively, from the mean position. Its time period is `underlinebb(2pisqrt((x_1^2 - x_2^2)/(v_2^2 - v_1^2)))`.

Explanation:

As we know in SHM,

v² = ω²(a² - x²)

⇒ `"v"_1^2 = omega^2(a^2 - x_1^2)` .......(i)

and `"v"_2^2 = omega^2(a^2 - x_2^2)` .........(ii)

Subtracting Eq. (i) from Eq. (ii), we get

`"v"_2^2 - "v"_1^2 = omega^2(x_1^2 - x_1^2)`

= `(4pi^2)/T^2(x_1^2 - x_2^2)` `(∵ omega = (2pi)/T)`

⇒ T = `2pisqrt((x_1^2 - x_2^2)/("v"_2^2 - "v"_1^2))`