Question

A block of mass 5 kg executes simple harmonic motion under the restoring force of a spring. The amplitude and the time period of the motion are 0.1 m and 3.14 s respectively. Find the maximum force exerted by the spring on the block.

• A. 2N
• B. 1N
• C. 0.5N
• D. 20 N

Answer 1

The correct option is A 2N

Period of motion given is 3.14 s

We know $T=\frac{2\pi }{\omega }\Rightarrow \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.14}=2$

Amplitude given is 0.1 m The equation of the SHM could be

$X=0.1sin(2t+{\varphi }_0)$

Where ${\varphi }_0$ could be the initial phase.
Now differentiating x with respect to time

$\frac{dx}{dt}=0.2cos(2t+{\varphi }_0)=v$

Differentiating again $\frac{d^{2}x}{d{t}^2}=-0.4sin(2t+{\varphi }_0)=a$

So acceleration is maximum when sin $(2t+{\psi }_0)=-1$

$a_{max}=0.4m{s}^{-2}$

Given mass = 5kg

And like Newton told us F = ma

So $F_{max}=m{a}_{max}$

$\Rightarrow F_{max}=5\times 0.4=2N$