In the arrangement shown in the diagram, pulleys are small and springs are ideal. k1=k2=k3=k4=10 N m- 1 are force constants of the springs and mass m=10 kg . If the time period of small vertical oscillations of the block of mass m is given by 2π textx seconds , then find the value of x .

Question

In the arrangement shown in the diagram, pulleys are small and springs are ideal. k1=k2=k3=k4=10 N m- 1 are force constants of the springs and mass m=10 kg . If the time period of small vertical oscillations of the block of mass m is given by 2π textx seconds , then find the value of x . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-24bnuctadupw.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

textitT=2π √( textitm/ textitK texteq) where textitK texteq=(1/4 [(1) textitk1 + (1) textitk2 + (1/ textitk3) + (1/ textitk4)]=(1 0/1 6) textit/T)=2π √( textitm/ textitK texteq)=2π textx So, textx = √(1 0/1 0 / 1 6) = 4 text, textx = 4

Q. In the arrangement shown in the diagram, pulleys are small and springs are ideal. k1​=k2​=k3​=k4​=10Nm−1 are force constants of the springs and mass m=10kg . If the time period of small vertical oscillations of the block of mass m is given by 2πx seconds , then find the value of x .

T=2πKeq​m​​ where Keq​=4[k1​1​+k2​1​+k3​1​+k4​1​]1​=1610​ T=2πKeq​m​​=2πx So, x=10/1610​​=4, x=4

$T = 2 π \frac{m}{K _{eq}}$
where $K_{eq} = \frac{1}{4 [ \frac{1}{k _{1}} + \frac{1}{k _{2}} + \frac{1}{k _{3}} + \frac{1}{k _{4}} ]} = \frac{10}{16}$
$T = 2 π \frac{m}{K _{eq}} = 2 π x$
So, $x = \frac{10}{10/16} = 4, x = 4$