SHMHard
Question
Vertical displacement of a plank with a body of mass ′m′ on it is verifying according to law y = sin ωt + √3 cosωt. The minimum value of ω for which the mass just breaks off the plank and the moment it occurs first after t = 0 are by : (y is positive vertically upwards)
Options
A.
B.
C.
D.
Solution
y = sinωt + √3 cosωt = 2sin 
To berks off mg
mg = mω2min A ⇒ g = 2ω2min ⇒ ω =
Moment it occurs first after t = 0
2 = 2sin
⇒ ωt1 = 
⇒ t1 = 
To berks off mg
mg = mω2min A ⇒ g = 2ω2min ⇒ ω =
Moment it occurs first after t = 0
2 = 2sin
⇒ ωt1 = ⇒ t1 = Create a free account to view solution
View Solution FreeMore SHM Questions
Two particle executes S.H.M of same amplitude and frequency along the same straight line. They pass one another when goi...A simple pendulum of length L has an energy E and amplitude A. The energies of the simple pendulum (i) when the length i...Which of the following statement is incorrect for an object executing S.H.M. :...The total mechanical energy of a particle of mass m executing SHM is E = 1/2mω2A2. If the particle is replaced by a...In question 35 which of the graph between velocity and time is correct ?...