A block of a mass 2 kg is attached with two identical springs of spring constant 20 N / m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is (π/√x) in SI unit. The value of x is

Question

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/705ec35d09539e74e6a5a8c951a77dbf-.png" /> A block of a mass 2 kg is attached with two identical springs of spring constant 20 N / m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is (π/√x) in SI unit. The value of x is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

F =-2 kx , a =-(2 kx / m ), ω=√(2 k / m )=√(2 × 20/2) =√20 rad / s T =(2 π/ω)=(2 π/√20)=(π/√5) x =5

F=−2kx,a=−m2kx​,ω=m2k​​=22×20​​ =20​rad/s T=ω2π​=20​2π​=5​π​ x=5

$F = - 2 k x, a = - \frac{2 k x}{m}, ω = \frac{2 k}{m} = \frac{2 \times 20}{2}$
$= 20 r a d / s$
$T = \frac{2 π}{ω} = \frac{2 π}{20} = \frac{π}{5}$
$x = 5$