Vertical oscillations

Vertical displacement of a plank with a body of mass m m on it is varying according to the law y = sin ω t + 3 cos ω t y= \sin{\omega t} + \sqrt{3}\cos{\omega t} . The minimum value of ω \omega such that the mass breaks off the plank and the moment it occur first after t = 0 t=0 is given by (y is positive vertically upwards)

ω = a g b \omega = \dfrac{\sqrt{ag}}{\sqrt{b}} and t = t= z π 6 g \dfrac{\sqrt{z}\pi}{6\sqrt{g}} respectevly. Find the value of a + b + z a+b+z

Check your Calibre

HERE a and b are positive coprime integers which are square free

The answer is 5.

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1 solution

Steven Chase
Mar 6, 2017

y = ω 2 y = ω 2 [ s i n ω t + 3 c o s ω t ] \large{y^{''} = -\omega^2 y = -\omega^2 [sin\omega t + \sqrt{3} cos\omega t]}

Find max magnitude of y y^{''} and equate it to g g

y m a x = ω 2 1 + 3 = 2 ω 2 = g ω m i n = g 2 \large{y^{''}_{max} = \omega^2 \sqrt{1 + 3} = 2 \omega^2 = g \\ \boxed{\omega_{min} = \frac{\sqrt{g}}{\sqrt{2}}}}

The block can only leave the plank traveling downward. Set y y^{''} equal to g -g and find the associated time.

g = g 2 [ s i n ω t l e a v e + 3 c o s ω t l e a v e ] s i n ω t l e a v e + 3 c o s ω t l e a v e = 2 ω t l e a v e = π 6 = g 2 t l e a v e t l e a v e = 2 π 6 g \large{-g = -\frac{g}{2} [sin \, \omega t_{leave} + \sqrt{3} cos \, \omega t_{leave}] \\ sin\,\omega t_{leave} + \sqrt{3} cos\,\omega t_{leave} = 2 \\ \implies \omega t_{leave} = \frac{\pi}{6}} = \frac{\sqrt{g}}{\sqrt{2}} t_{leave} \\ \boxed{\large{t_{leave} = \frac{\sqrt{2} \pi}{6 \sqrt{g}}}}

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