Right choice is (b) 67.2
To explain I would say: 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.
Maximum acceleration is
Amax = ω^2r = (5.7)^2 × 2.07 = 32.49 × 2.07
= 67.25 m/s^2.