Question

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 1

Explanation:

f1=6Hz

f2=8Hz

F = m r w² = m *r * 4 * pi² * f²

Hence F is proportional to f²

Now

f² = f1² + f2²

f² = 6² + 8² = 100

f = 10 hz

Answer 2

★$\large{\underline{\underline{\mathbf{Answer-}}}}$

✍ 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

$\large\bold{n = \frac{1}{2\pi} \sqrt{ \frac{ k_{1} + k_{2} }{m} } }$

$\large\bold{ = > n = \frac{1}{2\pi} \sqrt{ \frac{ {4\pi}^{2} { n_{1}}^{2} m + {4\pi}^{2} { n_{2} }^{2} m }{m} } }$

$\large\bold{ = > \sqrt{ { n_{1} }^{2} + { n_{2}}^{2} } }$

$\large\bold{ = > \sqrt{ {8}^{2} + {6}^{2} } }$

$\large\boxed{ = > 10Hz}$

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