A body executes simple harmonic motion under | vedclass

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(A) The force in $SHM$ is given by $F = -kx = -m\omega^2 x$. For force $F_1$,$k_1 = m\omega_1^2 = m(2\pi/T_1)^2$. For force $F_2$,$k_2 = m\omega_2^2 =

Vedclass · Accepted Answer

$12/25$

Vedclass · Answer

(A) The force in $SHM$ is given by $F = -kx = -m\omega^2 x$. For force $F_1$,$k_1 = m\omega_1^2 = m(2\pi/T_1)^2$. For force $F_2$,$k_2 = m\omega_2^2 = m(2\pi/T_2)^2$. When both forces act simultaneously,the resultant force is $F = F_1 + F_2 = -(k_1 + k_2)x$. Thus,the effective spring constant is $k = k_1 + k_2$. Substituting $k = m\omega^2$,we get $m\omega^2 = m\omega_1^2 + m\omega_2^2$,which simplifies to $\omega^2 = \omega_1^2 + \omega_2^2$. Using $\omega = 2\pi/T$,we have $(2\pi/T)^2 = (2\pi/T_1)^2 + (2\pi/T_2)^2$,which gives $1/T^2 = 1/T_1^2 + 1/T_2^2$. Solving for $T$,we get $T = \frac{T_1 T_2}{\sqrt{T_1^2 + T_2^2}}$. Given $T_1 = 4/5 \ s$ and $T_2 = 3/5 \ s$: $T = \frac{(4/5)(3/5)}{\sqrt{(4/5)^2 + (3/5)^2}} = \frac{12/25}{\sqrt{16/25 + 9/25}} = \frac{12/25}{\sqrt{25/25}} = \frac{12}{25} \ s$.

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