Question

A plank with a small block on top of it is undergoing vertical SHM. Its period is 2sec.The minimum amplitude at which the block will seperate from plank is?

Answer 1

Hello there,


● Answer -
A = 99.27 cm


● Explaination -
# Given -
T = 2 s

# Solution -
Let the block be in SHM of amplitude A and angular velocity ω.

Block will fall of the plank when a >= g,
a = g
Aω^2 = g
A = g / ω^2
A = gT^2 / 4π^2
A = 9.8 × 2^2 / 4×π^2
A = 0.9927 m
A = 99.27 cm

Therefore, minimum amplitude of SHM for block to fall off the plank is 99.27 cm.

Hope this helped you...

Answer 2

max acelration should be greater than or equal to g(9.8m/s^2) :》

• a max = g
• A w^2 = g
• A = g/w^2
• A=$\frac{g}{{\pi}^{2} }$
• {T=2${\pi}$/w 》 2=2${\pi}$/w 》 w=${\pi}$}