QJEEM23Ph0f74: JEE Main 2023 Physics - Class 11 JEEM23Ph0f74 Solution: A rectangular block of mass attached to a horizontal spiral…

To find the maximum force exerted by the spring on the block, we can use Hooke's law and the properties of simple harmonic motion.

First, let's find the angular frequency $\omega$ :

$\omega = \frac{2\pi}{T}$

where $T = 3.14\,\mathrm{s}$ is the time period.

$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$

Now, let's find the maximum velocity $v_{max}$ of the block:

$v_{max} = \omega A$

where $A = 1\,\mathrm{m}$ is the amplitude.

$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$

Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$ :

$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$

$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$

Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke's law:

$F_{max} = kA$

$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$

The maximum force exerted by the spring on the block is $20\,\mathrm{N}$ .