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 force constant 4 N/m and 16 N/m respectively. If T 1 and T 2 are the period of oscillatio

Question

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 force constant 4 N/m and 16 N/m respectively. If T 1 ​ and T 2 ​ are the period of oscillations of springs A and B respectively, then ( T 1 ​ + T 2 ​ ) / ( T 1 ​ − T 2 ​ ) will be

TestHub · Accepted Answer

The time period of oscillation of a spring-mass system is given by $T = 2 π \frac{m}{k}$ .

For spring A, $T_{1} = 2 π \frac{m}{k _{1}}$

For spring B, $T_{2} = 2 π \frac{m}{k _{2}}$

Taking the ratio of the time periods, we get:

$\frac{T _{1}}{T _{2}} = \frac{k _{2}}{k _{1}}$

Given $k_{1} = 4$ N/m and $k_{2} = 16$ N/m:

$\frac{T _{1}}{T _{2}} = \frac{16}{4} = 2$

This implies $T_{1} = 2 T_{2}$ .

Now, evaluating the required expression:

$\frac{T _{1} + T _{2}}{T _{1} - T _{2}} = \frac{2 T _{2} + T _{2}}{2 T _{2} - T _{2}} = \frac{3 T _{2}}{T _{2}} = \frac{3}{1}$

Thus, the ratio is $3 : 1$ .

Answer: $3 : 1$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion