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(A) For a body oscillating under a restoring force $F$,we have $F = kx$,where $k = m\omega^2$. Thus,$\omega^2 = \frac{F}{mx}$. For the first force $F_

Vedclass · Accepted Answer

$T=\sqrt{\frac{T_1^2 T_2^2}{T_1^2+T_2^2}}$

Vedclass · Answer

(A) For a body oscillating under a restoring force $F$,we have $F = kx$,where $k = m\omega^2$. Thus,$\omega^2 = \frac{F}{mx}$. For the first force $F_1$,$\omega_1^2 = \frac{F_1}{mx}$. For the second force $F_2$,$\omega_2^2 = \frac{F_2}{mx}$. When both forces act simultaneously,the resultant force is $F_{res} = F_1 + F_2$. The resultant angular frequency $\omega$ is given by $\omega^2 = \frac{F_1 + F_2}{mx} = \frac{F_1}{mx} + \frac{F_2}{mx} = \omega_1^2 + \omega_2^2$. Since $\omega = \frac{2\pi}{T}$,we have $\left(\frac{2\pi}{T}\right)^2 = \left(\frac{2\pi}{T_1}\right)^2 + \left(\frac{2\pi}{T_2}\right)^2$. Dividing by $4\pi^2$,we get $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$. $\frac{1}{T^2} = \frac{T_1^2 + T_2^2}{T_1^2 T_2^2}$. Therefore,$T = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}}$.

Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

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