A mass m is suspended separately by two different springs of spring constant K1 and K2 gives the time-period t1 and t2 respectively. If same mass m is connected by both springs as shown in figure then time-period t is given by the relation

Question

A mass m is suspended separately by two different springs of spring constant K1 and K2 gives the time-period t1 and t2 respectively. If same mass m is connected by both springs as shown in figure then time-period t is given by the relation <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-v3dy2uaqzy2i0ybh.jpg" /> (A) t=t1+t2 (B) t=(t1 ⋅ t2/t1 + t2) (C) t2=t12+t22 (D) t- 2=t1- 2+t2- 2

Tardigrade Admin · Accepted Answer

t1=2π √(m/K1) and t2=2π √(m/K2) Equivalent spring constant for shown combination is K1+K2. So time period t is given by t=2π √(m/K1 + K2) By solving these equations we get t- 2=t1- 2+t2- 2

Q. A mass $m$ is suspended separately by two different springs of spring constant $K_{1}$ and $K_{2}$ gives the time-period $t_{1}$ and $t_{2}$ respectively. If same mass $m$ is connected by both springs as shown in figure then time-period $t$ is given by the relation

$t = t_{1} + t_{2}$

$t = \frac{t _{1} \cdot t _{2}}{t _{1} + t _{2}}$

$t^{2} = t_{1}^{2} + t_{2}^{2}$

$t^{- 2} = t_{1}^{- 2} + t_{2}^{- 2}$

$t_{1} = 2 π \frac{m}{K _{1}}$ and $t_{2} = 2 π \frac{m}{K _{2}}$
Equivalent spring constant for shown combination is
$K_{1} + K_{2} .$ So time period $t$ is given by $t = 2 π \frac{m}{K _{1} + K _{2}}$
By solving these equations we get $t^{- 2} = t_{1}^{- 2} + t_{2}^{- 2}$

Q. A mass m is suspended separately by two different springs of spring constant K1​ and K2​ gives the time-period t1​ and t2​ respectively. If same mass m is connected by both springs as shown in figure then time-period t is given by the relation

t=t1​+t2​

t=t1​+t2​t1​⋅t2​​

t2=t12​+t22​

t−2=t1−2​+t2−2​