Right answer is (a) 1.1

Explanation: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation

v = \(\omega \sqrt{r^2 – x^2}\)

Given values:

when point (x) = 0.75 m, then its velocity (v) = 11 m/s

when point (x) = 2 m, then its velocity (v) = 3 m/s

When point is 0.75 m away from the mid path (v) is,

v = \(\omega \sqrt{r^2 – x^2}\)

11 = \(\omega \sqrt{r^2 – (0.75)^2}\) —>Eq(1)

Similarly, When point is 2 m away from the centre (v) is,

v = \(\omega \sqrt{r^2 – x^2}\)

3 = \(\omega \sqrt{r^2 – 2^2}\) —>Eq(2)

Solving Eq(1) & Eq(2) we get,

\(\frac{11}{3} \frac{\omega \sqrt{r^2 – (0.75)^2}}{\omega \sqrt{r^2 – 2^2}} = \frac{\sqrt{r^2 – (0.75)^2}}{\sqrt{r^2 – 2^2}} \)

Squaring on both sides, we get

\(\frac{121}{9} = \frac{r^2 – 0.5625}{r^2 – 4}\)

121r^2 – 484 = 9r^2 – 5.0625

121r^2 – 9r^2 = 484 – 5.0625

112r^2 = 478.9375

r^2 = \(\frac{478.9375}{112}\)

r^2 = 4.27622

r = 2.07m

Substituting the value of r in Eq(1) we get,

11 = \( \omega \sqrt{(2.07)^2 – (0.75)^2}\)

11 = \( \omega \sqrt{4.2849 – 0.5625}\)

11 = \( \omega \sqrt{3.7224}\)

11 = 1.9293 ω

ω = \(\frac{11}{1.9293}\) = 5.7 rad/s.

We know periodic time is,

Tp = \(\frac{2 \pi}{\omega} = \frac{2 \pi}{5.7} = \frac{2 \times 3.14}{5.7} = \frac{6.28}{5.7}\) = 1.1 s.