Question

A block of mass $m$ suspended from a spring executes vertical SHM of time period $T$ as shown in the figure. The amplitude of the SHM is $A$ and the spring is never in compressed state during the oscillation. The magnitude of minimum force exerted by the spring on the block is

• A. $mg-\frac{4{\pi }^2}{T^2}mA$
• B. $mg+\frac{4{\pi }^2}{T^2}mA$
• C. $mg-\frac{{\pi }^2}{T^2}mA$
• D. $mg+\frac{{\pi }^2}{T^2}mA$

Answer 1

The correct option is A $mg-\frac{4{\pi }^2}{T^2}mA$


The spring is never compressed. Hence, spring exerts least force on the block when the block is at topmost position during SHM.
At equilibrium position, we know that $k{x}_0=mg\Rightarrow x_0=\frac{mg}{k}$
$F_{\text{min}}=k(x_0-A)=k[\frac{mg}{k}-A]$
$\Rightarrow F=mg-m{\omega }^{2}A$
[$\because {\omega }^2=\frac{k}{m}$]
Using $\omega =\frac{2\pi }{T}$, we get
$F_{min}=mg-\frac{4{\pi }^2}{T^2}mA$
Thus, option (a) is the correct answer.