- What is the frequency of the motion?
- What is the amplitude of the motions?
- What is the forces constant of the spring?

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Solution:

Here is given, block of mass (m) = 250gm = 250 x 10^{-3} kg

Time (t) = 0.15 sec

Displacement (x) = 40cm = 40 x 10^{-2}m

Frequency (f) = ?

Force constant (k) = ?

We know,

**a.** f = 1/t = 1/0.15 = 6.67 rev sec^{-1}

**b.** Amplitude (x_{m}) =

\(\frac{40 \times 10^{-2}}{2}\) = 20 x 10^{-2} m

**c. **Again we know that

\(2\pi f = \sqrt{\frac{k}{m}}\)

\(4\pi^2f^2 = \frac{k}{m}\)

\(k = 4\pi^2f^2 m\)

\(k = 4\pi^{2}(6.67)^2 \times 250 \times 10^{-3}\)

k = 439.08 Nm^{-1}

Hence, required frequency, amplitude and force constant are 6.67 rev s^{-1}, 20 x 10^{-2}m and 439.08 Nm^{-1}

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