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Step-by-Step Solution
Step 1: Express the force in terms of displacement
The given periodic force is
F = -25\,x\,(\text{N}).
The mass of the particle is
m = 250\,\text{g} = 0.25\,\text{kg}.
By Newton’s second law, acceleration a is
a = \frac{F}{m}.
Substituting F = -25\,x and m = 0.25\,\text{kg} :
a = \frac{-25\,x}{0.25} = -100\,x.
Step 2: Compare with standard SHM form
In simple harmonic motion (SHM), acceleration is given by
a = -\omega^2\,x.
Comparing -100\,x with -\omega^2\,x shows that
\omega^2 = 100,
thus
\omega = 10\,\text{rad/s}.
Step 3: Relation for maximum velocity in SHM
The maximum velocity v_{\max} in SHM is
v_{\max} = \omega\,A,
where A is the amplitude. We are told the maximum speed is
4\,\text{m/s}.
So:
4 = 10 \times A.
Step 4: Solve for the amplitude
From
4 = 10 \times A,
we get
A = \frac{4}{10} = 0.4\,\text{m}.
Converting to centimeters:
A = 0.4\,\text{m} \times 100\,\frac{\text{cm}}{\text{m}} = 40\,\text{cm}.
Final Answer
The amplitude of the motion is 40\,\text{cm}.
