Simple Pendulum Frequency Calculation: Acceleration and Homework Equations

AI Thread Summary
The frequency of a simple pendulum is affected by the acceleration of its support. When the support accelerates upward at 0.50g, the effective gravitational force increases to 1.50g. The formula for frequency, f = 1/2π√(g/L), requires this adjusted value of g. The discussion highlights that the perceived increase in gravity results from the additional force acting on the pendulum due to the upward acceleration. Understanding this concept is crucial for accurately calculating the pendulum's frequency in an accelerating frame of reference.
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Homework Statement



A simple pendulum oscillates with frequency f. What is its frequency if it accelerates at 0.50g upward


Homework Equations



f = 1/2pi sqrt(g/L)


The Attempt at a Solution



I want to say that g in the equation is equal to 0.50g, but I know this is wrong. Why do I plug in a value of 1.50g for g?
 
Physics news on Phys.org
If you are in a lift that starts to accelerate upward your feet would feel an increased push. When you stand on a train that starts to accelerate forward, you feel a force pushing you backwards. This is because of the law of inertia, Newton's first law. Every body stays in rest or keeps its initial velocity until the action of a force. If the lift or train accelerates, you need an extra force to accelerate together with it, and you feel it as a force equal to -ma, the product of your mass to the acceleration of your system, and pointing opposite direction as the acceleration.

The pendulum hangs from a support that starts to accelerate with 0.5 g. It is the same as if an extra force 0.5 mg acted on it in addition to the original mg force in the accelerating frame of reference, that is an increased gravity.

ehild
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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