SHMHard
Question
The speed v of a particle moving along a straight line, when it is at a distance (x) from a fixed point of the line is given by
v2 = 108 − 9x2 (assuming mean position to have zero phase constant)
(all quantities are in cgs units) :
v2 = 108 − 9x2 (assuming mean position to have zero phase constant)
(all quantities are in cgs units) :
Options
A.the motion is uniformly accelerated along the straight line
B.the magnitude of the acceleration at a distance 3cm from the fixed point is 27 cm/s2
C.the motion is simple harmonic about the given fixed point.
D.the maximum displacement from the fixed point is 4 cm.
Solution
v2 = 108 − 9x2
= − 18 x ⇒ acc. A = −9x (non-uniform)
at x = 3cm
a = −27 or |a| = 27cm/s2
also a = − 9x is a S.H.M equation so particle perform S.H.M about the give fixed point
V is maximum at x = 0 V, x = 0
and V is Zero at x =
So Amplitude = 2√3cm
= − 18 x ⇒ acc. A = −9x (non-uniform) at x = 3cm
a = −27 or |a| = 27cm/s2
also a = − 9x is a S.H.M equation so particle perform S.H.M about the give fixed point
V is maximum at x = 0 V, x = 0
and V is Zero at x =
So Amplitude = 2√3cm
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