Question

A particle executing SHM of amplitude 5 cm and has acceleration of 27cm/sec2 when it is 3cm from the mean position on it's maximum velocity is
a) 15cm/sec
b) 30 cm/sec
c) 45 CM/sec
d) 60 cm/sec​

Answer 1

Dear Student,

◆ Answer -

v = 15 cm/s

● Explanation -

# Given -

A = 5 cm

a = 27 cm/s^2

x = 3 cm

# Solution -

Acceleration is calculated by -

a = xw^2

27 = 3 × w^2

w^2 = 9

w = 3 rad/s

Maximum velocity of the particle in SHM is -

v = Aw

v = 5 × 3

v = 15 cm/s

Hence, maximum velocity of the particle is 15 cm/s.

Thanks dear. Hope this helps you...

Answer 2

A is the correct answer

______________________________

A = 5

a = 27cm/sec^2

x = 3cm

Formula = a = xw^2

a = 27,x = 3,so w = ? ,^2

27 = 3 × w ^2

w^2 = 9

w = 3 rad/s

$v = aw \\ = 5 \times 3 = 15$

#AnswerWithQuality

#BAL