Given,
v = 10 cm/sec
amax = 50 cm/s2
v = rω
10 = rω
r2ω2 = 100
ω2 = \(\frac{100}{r^2}\) ....(1)
Maximum acceleration,
amax = ω2r
50 = ω2r
ω2 = \(\frac{50}{r}\) ....(2)
Evaluate equation (1) and (2),
\(\frac{100}{r^2}\) = \(\frac{50}{r}\)
r = 2 cm
Then,
ω2 = \(\frac{100}{r^2}\)
ω = \(\sqrt{\frac{100}{r^2}}\)
∵ r = 2
ω = \(\sqrt{\frac{100}{4}}\)
ω = \(\sqrt{25}\)
ω = 5 sec2
The speed of particle at distance of 1 cm from the mean position,
v2 = ω2(r2 - v2)
v2 = (5)2[(2)2-(1)2]
v2 = 25[4-1]
v2 = 75
v2 = 75
v = 5√3 cm/sec
∴ Option (B) is correct.
