When a body is suspended from a fixed point by a spring, the angular frequency of its vertical oscillations is ω_1 . When a different spring is used, the angular frequency is ω_2 . The angular frequency of vertical

Correct answer: C. ω=[ω_1^2 ω_2^2 /(ω_1^2+ω_2^2)]^ 1 / 2 No Solution

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When a body is suspended from a fixed point by a spring, the...

When a body is suspended from a fixed point by a spring, the angular frequency of its vertical oscillations is

\omega_1

. When a different spring is used, the angular frequency is

\omega_2

. The angular frequency of vertical oscillations when both the springs are used together in series is given by

A $\omega=\left[\omega_1^2+\omega_2^2\right]^{1 / 2}$
B $\omega=\left[\left(\omega_1^2+\omega_2^2\right) / 2\right]^{1 / 2}$
C $\omega=\left[\omega_1^2 \omega_2^2 /\left(\omega_1^2+\omega_2^2\right)\right]^{1 / 2}$
D $\omega=\left[\omega_1^2 \omega_2^2 / 2\left(\omega_1^2+\omega_2^2\right)\right]^{1 / 2}$

C. $\omega=\left[\omega_1^2 \omega_2^2 /\left(\omega_1^2+\omega_2^2\right)\right]^{1 / 2}$

No Solution