Question 282: Oscillations

Question

Under the influence of force $F_1$ the body oscillates with a period $T_1$ and due to another force $F_2$ body oscillates with period $T_2$. If both forces acts simultaneously, then the resultant period is
(consider displacement is same in all three cases)

Accepted Answer

Time period for the force $F_1=T_1$
Time period for the force $F_2=T_2$
If both forces acts simultaneously then the resultant force is $F$,
$F=F_1+F_2$
Let, time period for the force $F$ be $T$
Now, $F=F_1+F_2$
$\begin{aligned}
& m \omega^2 y=m \omega_1^2 y+m \omega_2^2 y \quad\left(\because F=m \omega^2 yight) \
& \omega^2=\omega_1^2+\omega_2^2
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad\left(\frac{2 \pi}{T}ight)^2=\left(\frac{2 \pi}{T_1}ight)^2+\left(\frac{2 \pi}{T_2}ight)^2 \
& \Rightarrow \quad \frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2} \
& \Rightarrow \quad \frac{1}{T^2}=\frac{T_2^2+T_1^2}{T_1^2 T_2^2} \
& \Rightarrow \quad T^2=\frac{T_1^2 T_2^2}{T_1^2+T_2^2} \Rightarrow T=\sqrt{\frac{T_1^2 T_2^2}{T_1^2+T_2^2}} \
&
\end{aligned}$

Answer

$T=\sqrt{\frac{T_1^2+T_2^2}{T_1^2 T_2^2}}$

Answer

$T=\sqrt{\frac{T_1^2 T_2^2}{T_1^2+T_2^2}}$

Answer

$T=\sqrt{\frac{T_1^2}{T_2^2}}$

Answer

$T=\sqrt{T_1^2+T_2^2}$

Oscillations

Under the influence of force F_1 the body oscillates with a period T_1 and due to another force F_2 body oscillates with period T_2. If both forces acts simultaneously, then the resultant period is (consider displacement is same in all three cases)

Choose the correct answer: