SHMHard
Question
A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest. Then :
Options
A.the period of oscillation is 
B.the block weighs double its weight, when the plank is at one of the positions of momentary rest.
C.the block weighs 0.5 times its weight on the plank halfway up
D.the block weighs 1.5 times its weight on the plank halfway down
Solution
Given A = 0.4m , and a = g
so ω2A = g ⇒ ω2 =
= 25
⇒ ω = 5 T =
= 2π/5 sec.
At lowest position acceleration. = ω2A + g = g + g = 2g
So weight = m (2g) = 2mg
at half distance a = g/2
So weight at upper half distance = m(g − g/2) = mg/2
and weight at lower half distance
= m(g + g/2) =
actual weight at equilibrium position (maximum v)
so ω2A = g ⇒ ω2 =
= 25⇒ ω = 5 T =
= 2π/5 sec.At lowest position acceleration. = ω2A + g = g + g = 2g
So weight = m (2g) = 2mg
at half distance a = g/2
So weight at upper half distance = m(g − g/2) = mg/2
and weight at lower half distance
= m(g + g/2) =
actual weight at equilibrium position (maximum v)
Create a free account to view solution
View Solution FreeMore SHM Questions
A simple pendulum is suspended from the ceiling of a vehicle, its time period is T. Vehicle is moving with constant velo...The distance between the points of suspension and centre of oscillation for a compound pendulum is :...The distance of hinge point of a compound pendulum from its centre of gravity is l, the time period of oscillation relat...A toy car of mass m is having two similar rubber ribbons attached to it as shown in the figure. The force constant of ea...The bob in a simple pendulum of length l is released at t = 0 from the position of small angular displacement θ. Li...