Due to some force F_1,a body oscillates with | vedclass

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(C) Under the influence of force $F_1$,the angular frequency is $\omega_1$,where $F_1 = m\omega_1^2 y$. The period is $T_1 = 4/5 \, s$.Under the influ

Vedclass · Accepted Answer

$0.48$

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(C) Under the influence of force $F_1$,the angular frequency is $\omega_1$,where $F_1 = m\omega_1^2 y$. The period is $T_1 = 4/5 \, s$.Under the influence of force $F_2$,the angular frequency is $\omega_2$,where $F_2 = m\omega_2^2 y$. The period is $T_2 = 3/5 \, s$.When both forces act simultaneously,the net force is $F = F_1 + F_2 = m\omega^2 y$.Thus,$m\omega^2 y = m\omega_1^2 y + m\omega_2^2 y$,which implies $\omega^2 = \omega_1^2 + \omega_2^2$.Substituting $\omega = 2\pi/T$,we get $(2\pi/T)^2 = (2\pi/T_1)^2 + (2\pi/T_2)^2$.This simplifies to $1/T^2 = 1/T_1^2 + 1/T_2^2$,or $T = \sqrt{\frac{T_1^2 T_2^2}{T_1^2 + T_2^2}}$.Substituting the values: $T = \sqrt{\frac{(4/5)^2 	imes (3/5)^2}{(4/5)^2 + (3/5)^2}} = \sqrt{\frac{(16/25) 	imes (9/25)}{16/25 + 9/25}} = \sqrt{\frac{144/625}{25/25}} = \sqrt{\frac{144}{625}} = \frac{12}{25} = 0.48 \, s$.

Column-$I$	Column-$II$
$(A)$ Velocity-displacement graph $(\omega \neq 1)$	$(i)$ Straight line
$(B)$ Acceleration-displacement graph	$(ii)$ Sinusoidal
$(C)$ Acceleration-time graph	$(iii)$ Circle
$(D)$ Acceleration-velocity graph $(\omega \neq 1)$	$(iv)$ Ellipse

Due to some force $F_1$,a body oscillates with a period of $4/5 \, s$,and due to another force $F_2$,it oscillates with a period of $3/5 \, s$. If both forces act simultaneously,the new period will be .... $s$.

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