SHMMedium
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 constant with the plank when latter is at momentary rest. Then :
Options
A.the period of oscillation is 
B.the block weighs double of its weight, when the plank is at one of the positions of momentary rest
C.the block weighs 0.5 times on the plank halfway up
D.the plank block weights 1.5 time its weight on the planks halfway down
Solution
The position of momentary rest in S.H.M is extreme position where velocity of particle is zero
As the block less contact with the plank at this position i.e. normal force becomes zero, it has to be the upper extremes where acceleration of the block will be g downwards.
∴ ω2A = g ⇒ ω2 =
= 25 = 5 red /s
Therefore period =
s
Acceleration in S.H.M. is given by a = ω2x
From the figure we can see that the , At lower extreme, acceleration is g upwards
∴ N - mg = ma ⇒ N = m(a + g) = 2mg
At halfway up, acceleration is g/2 down wards
∴ mg - N = ma ⇒ N = m (g -
) = mg
As the block less contact with the plank at this position i.e. normal force becomes zero, it has to be the upper extremes where acceleration of the block will be g downwards.
∴ ω2A = g ⇒ ω2 =
= 25 = 5 red /sTherefore period =
sAcceleration in S.H.M. is given by a = ω2x
From the figure we can see that the , At lower extreme, acceleration is g upwards
∴ N - mg = ma ⇒ N = m(a + g) = 2mg
At halfway up, acceleration is g/2 down wards
∴ mg - N = ma ⇒ N = m (g -
) = mg Create a free account to view solution
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