frequency of oscillation of a body is 6Hz when force F1 is applied and

1.	frequency of oscillation of a body is 6Hz when force F1 is applied and 8Hz when F2 is applied. if both forces F1 and F2 are applied together then find out the frequency of oscillation ?
Answer» ★ $\large{\underline{\underline{\mathbf{Answer-}}}}$ ✍ 10Hz ★ $\large{\underline{\underline{\mathbf{Explanation-}}}}$ $\star{\mathtt{\blue{\underline{Given:-}}}}$ Frequency of 1ST FORCE = 6Hz Frequency of 2ND force = 8Hz $\star{\mathtt{\blue{\underline{To\:find:-}}}}$ Frequency of oscillation when $\bold{ F_{1} \: and \: F_{2}}$ are APPLIED together. $\star{\mathtt{\blue{\underline{Solution-}}}}$ $\bold{According \:to\:question,}$ $\large\bold{ F_{1} = - k_{1}x}$ and, $\large\bold{ F_{2} = - k_{2}x}$ so, $\large\bold{ n_{1} = \frac{1}{2\pi} \sqrt{ \frac{ k_{1}}{m} } }$ $\large\bold{ = > 6Hz}$ $\large\bold{ n_{2} = \frac{1}{2\pi} \sqrt{ \frac{ k_{2}}{m} } }$ $\large\bold{ = > 8hz}$ Now, $\large\bold{F = F_{1} + F_{2}}$ $\large\bold{ = > - ( k_{1} + k_{2})x}$ THEREFORE $\large\bold{n = \frac{1}{2\pi} \sqrt{ \frac{ k_{1} + k_{2} }{m} } }$ $\large\bold{ = > n = \frac{1}{2\pi} \sqrt{ \frac{ {4\pi}^{2} { n_{1}}^{2} m + {4\pi}^{2} { n_{2} }^{2} m }{m} } }$ $\large\bold{ = > \sqrt{ { n_{1} }^{2} + { n_{2}}^{2} } }$ $\large\bold{ = > \sqrt{ {8}^{2} + {6}^{2} } }$ $\large\boxed{ = > 10Hz}$ __________________________________