A mass ' m 'is suspended from two coupled springs connected in series. The force constant for springs are k 1 and k 2 . The time period of the suspended mass will be

Question

A mass ' m 'is suspended from two coupled springs connected in series. The force constant for springs are k 1 and k 2 . The time period of the suspended mass will be (A) T =2 π √( m / k 1- k 2) (B) T =2 π √( mk 1 k 2/ k 1+ k 2) (C) T =2 π √( m / k 1+ k 2) (D) T =2 π √( m ( k 1+ k 2)/ k 1 k 2)

Tardigrade Admin · Accepted Answer

Ke q=(k1 k2/k1+k2) T=2 π √(m/ke q)

Q. A mass ' $m$ 'is suspended from two coupled springs connected in series. The force constant for springs are $k_{1}$ and $k_{2} .$ The time period of the suspended mass will be

$T = 2 π \frac{m}{k _{1} - k _{2}}$

$T = 2 π \frac{m k _{1} k _{2}}{k _{1} + k _{2}}$

$T = 2 π \frac{m}{k _{1} + k _{2}}$

$T = 2 π \frac{m ( k _{1} + k _{2} )}{k _{1} k _{2}}$

$K_{e q} = \frac{k _{1} k _{2}}{k _{1} + k _{2}}$
$T = 2 π \frac{m}{k _{e q}}$

Q. A mass ' m 'is suspended from two coupled springs connected in series. The force constant for springs are k1​ and k2​. The time period of the suspended mass will be

T=2πk1​−k2​m​​

T=2πk1​+k2​mk1​k2​​​

T=2πk1​+k2​m​​

T=2πk1​k2​m(k1​+k2​)​​