Question

A block of mass 0.5 kg hanging from a vertical spring executes simple harmonic motion of amplitude 0.1 m and time period 0.314 s. Find the maximum force exerted by the spring on the block.

Answer 1

Explanation:

• The maximum force exerted by the spring on the block can be calculated from the formula
• $F=\left(k^{*} x\right)+$ weight of the block where k is the spring constant and x is the amplitude.  It is given that $x=0.1 \mathrm{m}$,  $T=0.314 s$ and m = 0.5 kg.  We know that time period $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ and substituting the known values, we get spring constant $k=200 N / m$.
• Thus, from the formula $F=k * x$, we get $\mathrm{F}=2 \mathrm{ON}$. The maximum force exerted by the spring on the block is $F=25 N$.