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
The energy at the mean position of a pendulum will be :...A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic ...A particle executing SHM. Its time period is equal to the smallest time interval in which particle acquires a particular...A simple harmonic motion is given by y = 5 (sin 3π t + √3 cos 3π t). What is the amplitude of motion if ...A particle executes simple harmonic motion under the restoring force provided by a spring. The time period is T. If the ...