Question:-
What is the maximum acceleration of a cam follower undergoing simple harmonic motion?
Option (A)
(h / 2) (pw / Æ)2
Option (B)
2h (w2 / Æ2)
Option(C)
4h (w2 / Æ)
Option(D)
(2hpw2 / Æ2)
Correct Option:
(A)
Question Solution:

S = (h/2) – (h/2) cosb
= (h/2) (1-cosb)
Now (b/p) = (q/q0)
\ b = p(q/q0)
S = (h/2) [1-cos(pq / q0)]
w = (q/t); \ q = wt
S = (h/2) [1-cos(pwt / q0)]
Differentiating with respect to t; we get
V = (ds/dt) = {(h/2)(pw/q0)} sin (pwt / q0)
Maximum velocity
Vmax = (h/2)( pw / q0) at q = (q0/2)
Next, differentiating will respect to t; we get
f = (dV/dt) = (h/2) (pw/q0)2 cos (pwt / qt)
f = (h/2)(pw/q0)2 cos(pq / q0)
Maximum acceleration of follower
fmax = (h/2)(pw/q0)2 at q = 0
question-answer-faq-6373