Discussion Forum

A body of mass 64 g is made to oscillate turn by turn on two different springs A and B. spring A and B has a force constant 4$\frac{N}{m}$ and 16 $\frac{N}{m}$ respectively. If T₁ and T₂ are period of oscillations of spring A and B respectively , Then $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$ will be

Answer:

Explanation:

 Given  , m=64g g= 64 x 10^-3 kg

 k_A  =4 N/m and k_B⁼16 N/m

 The time period of oscillation of a spring

 $T=2\pi\sqrt{\frac{m}{k}}$

 $\Rightarrow T_{A}=T_{1}=2\pi\sqrt{\frac{m}{k_{A}}}=2\pi\sqrt{\frac{64\times 10^{-3}}{4}}$

                     =$2\pi\sqrt{16\times 10^{-3}}$  ...........(i)

 and  

and $T_{B}=T_{2}=2\pi\sqrt{\frac{m}{k_{B}}}=2\pi\sqrt{\frac{64\times 10^{-3}}{16}}$
$=2\pi\sqrt{4\times 10^{-3}}$..........(ii)

  Dividing Eq.(i) by Eq.(ii) , we get

$\frac{T_{1}}{T_{2}}=\sqrt{\frac{16\times 10^{-3} }{4\times 10^{-3}}}=2$

$\Rightarrow$     $T_{1}=2T_{2}$

 $\therefore$   $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}=\frac{2T_{2}+T_{2}}{2T_{2}-T_{2}}=\frac{3T_{2}}{T_{2}}=\frac{3}{1}or 3:1$