The maximum force acting on a particle executing simple harmonic motion is \(10 \mathrm{~N}\). The force on…
The maximum force acting on a particle executing simple harmonic motion is \(10 \mathrm{~N}\). The force on the particle when it is midway between mean and extreme positions will be
\(10 \mathrm{~N}\)
\(12 \mathrm{~N}\)
\(5 \mathrm{~N}\)
zero
Solution
Maximum force on the particle performing SHM, \(F_{\max }=10 \mathrm{~N}\)
We know that, In SHM, when body is at maximum displacement (amplitude \(a\)), then force on the particle is maximum.
\(\begin{aligned}
& \therefore \quad F_{\max }=\text { mass } \times \text { acceleration }=m \cdot \alpha_{\max } \quad {\left[\therefore \alpha_{\max }=\omega^2 a\right]} \\
& \Rightarrow \quad 10=m \cdot \omega^2 a \Rightarrow 10=m a \omega^2 \quad \ldots(\mathrm{i})
\end{aligned}\)
Force on the particle when it is mid way \(\left(y=\frac{a}{2}\right)\) between mean and extreme position,
\(\begin{aligned}
& F=m \omega^2 y=m \omega^2 \cdot\left(\frac{a}{2}\right) \quad\left[\because y=\frac{a}{2}\right] \\
& =\frac{\operatorname{ma\omega }^2}{2}=\frac{10}{2} \quad \text { [from Eq. (i)] } \\
& =5 \mathrm{~N} \\
\end{aligned}\)