A block of mass m hangs from three springs having same spring constant k. If the mass is slightly displaced downwards, the time period of oscillation will be

Question

A block of mass m hangs from three springs having same spring constant k. If the mass is slightly displaced downwards, the time period of oscillation will be (A) 2 π √(m/3 k) (B) 2 π √(3 m/2 k) (C) 2 π √(2 m/3 k) (D) 2 π √(3 k/m)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/67978b1581b9e91454d73650f6327697-.png" /> The first two springs are in parallel. So, k text eq of 1 text st 2 will be =2 k Then it becomes The springs 2 k and k are in series. So, k eq =(2 k × k/2 k+k) =(2 k × k/3 k)=(2/3) k T =2 π √(m/ke q) ⇒ T =2 π √(3 m/2 k)

Q. A block of mass $m$ hangs from three springs having same spring constant $k$ . If the mass is slightly displaced downwards, the time period of oscillation will be

$2 π \frac{m}{3 k}$

$2 π \frac{3 m}{2 k}$

$2 π \frac{2 m}{3 k}$

$2 π \frac{3 k}{m}$

The first two springs are in parallel.
So, $k_{eq}$ of $1^{st} 2$ will be $= 2 k$
Then it becomes
The springs $2 k$ and $k$ are in series.
So, $k_{e q} = \frac{2 k \times k}{2 k + k}$
$= \frac{2 k \times k}{3 k} = \frac{2}{3} k$
$T = 2 π \frac{m}{k _{e q}}$
$\Rightarrow T = 2 π \frac{3 m}{2 k}$

Q. A block of mass m hangs from three springs having same spring constant k. If the mass is slightly displaced downwards, the time period of oscillation will be

2π3km​​

2π2k3m​​

2π3k2m​​

2πm3k​​