Initially system is in equilibrium. Time period of SHM of block in vertical direction is:

Question

Initially system is in equilibrium. Time period of SHM of block in vertical direction is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-sstnwi7u6xiinxhi.jpg" /> (A) 2π √(m/3 k) (B) 2π √(m/2 k) (C) 2π √(m/k) (D) 2π √(2 m/k)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-sstnwi7u6xii2ctl.jpg" /> Net vertical force =kx+(k x/√2)cos45× 2=2kx=ma a=(2 k/m)x ⇒ ω =√(2 k/m) τ=2π √(m/2 k)

Q. Initially system is in equilibrium. Time period of SHM of block in vertical direction is :

$2 π \frac{m}{3 k}$

$2 π \frac{m}{2 k}$

$2 π \frac{m}{k}$

$2 π \frac{2 m}{k}$

Net vertical force $= k x + \frac{k x}{2} cos 45 \times 2 = 2 k x = ma$
$a = \frac{2 k}{m} x$
$\Rightarrow ω = \frac{2 k}{m}$
$τ = 2 π \frac{m}{2 k}$

Q. Initially system is in equilibrium. Time period of SHM of block in vertical direction is :

2π3km​​

2π2km​​

2πkm​​

2πk2m​​

Net vertical force =kx+2​kx​cos45×2=2kx=ma a=m2k​x ⇒ω=m2k​​ τ=2π2km​​

$2 π \frac{m}{3 k}$

$2 π \frac{m}{2 k}$

$2 π \frac{m}{k}$

$2 π \frac{2 m}{k}$

Net vertical force $= k x + \frac{k x}{2} cos 45 \times 2 = 2 k x = ma$
$a = \frac{2 k}{m} x$
$\Rightarrow ω = \frac{2 k}{m}$
$τ = 2 π \frac{m}{2 k}$