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

Question

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are k1 and k2. The time period of the suspended mass will be (A) T = 2 π √ ( m/ k1 - k2 )  (B) T = 2 π √ ( m k1 k2/ k1 + k2 )  (C) T = 2 π √ ( m/ k1 + k2 )  (D) T = 2 π √ ( m ( k1 + k2)/ k1k2 )

Tardigrade Admin · Accepted Answer

The effective spring constant of two springs in series is k=(k1 k2/k1+k2) Time period, T=2 π √(m/k)=2 π √(m(k1+k2)/k1 k2)

Q. A mass $m$ is suspended from the 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

$T = 2 π \frac{m}{k _{1} - k _{2}}$

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

$T = 2 π \frac{m}{k _{1} + k _{2}}$

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

The effective spring constant of two springs in series is $k = \frac{k _{1} k _{2}}{k _{1} + k _{2}}$
Time period, $T = 2 π \frac{m}{k} = 2 π \frac{m ( k _{1} + k _{2} )}{k _{1} k _{2}}$

Q. A mass m is suspended from the 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​)​​