A trolley is sliding down a frictionless slope having inclination θ. If a simple pendulum is sus

step 1 Hello and welcome to this video solution of numerate. Here we are given a trolley that is placed

A trolley is sliding down a frictionless slope having...

A trolley is sliding down a frictionless slope having inclination $\theta$. If a simple pendulum is suspended on top of this trolley, its periodic time is given by $\mathrm{T}=2 \pi \sqrt{\left(1 / \mathrm{g}_{\mathrm{eff}}\right)}$, where $g_{\mathrm{eff}}=\ldots \ldots$ (A) $\mathrm{g}$ (B) $g \sin \theta$ (C) $g \cos \theta$ (D) $g \tan \theta$

A trolley is sliding down a frictionless slope having inclination $\theta$. If a simple pendulum is suspended on top of this trolley, its periodic time is given by $\mathrm{T}=2 \pi \sqrt{\left(1 / \mathrm{g}_{\mathrm{eff}}\right)}$, where $g_{\mathrm{eff}}=\ldots \ldots$
(A) $\mathrm{g}$
(B) $g \sin \theta$
(C) $g \cos \theta$
(D) $g \tan \theta$

Key Concepts

Effective Gravitational Acceleration

The effective gravitational acceleration is the vector sum of the real gravitational acceleration and the pseudo acceleration due to the non-inertial frame’s motion. In the case of the trolley sliding down a slope, the component of the gravitational force perpendicular to the slope combines with the inertial effects, resulting in an effective acceleration of g cos? that governs the pendulum's oscillations.

Pseudo Force

In an accelerating frame, a pseudo force appears to act on all masses, and it is proportional to the acceleration of the frame but opposite in direction. For the accelerating trolley, this pseudo force combines with the gravitational force to alter the effective acceleration experienced by the pendulum, thereby modifying its equilibrium position and oscillatory behavior.

Simple Pendulum Period

The period of a simple pendulum, given by T = 2??(L/g_eff), represents the time it takes for the pendulum to complete one full oscillation. This period depends on the length of the pendulum and the effective gravitational acceleration acting on it, illustrating how gravity influences oscillatory motion.

Non-Inertial Reference Frames

When a system is accelerating, such as the trolley sliding down a frictionless slope, the frame of reference becomes non-inertial. In non-inertial frames, additional apparent forces, called pseudo forces, must be introduced to account for the acceleration when applying Newton’s laws, affecting the perceived behavior of objects like a pendulum.

Transcript

00:02 Hello and welcome to this video solution of numerate.

00:05 Here we are given a trolley that is placed at an angle theta in an entire plane and there is a simple pendulum that is suspended at the top of this trolley right so something let's say like this you've got a pendulum that is fixed to the trolley and this trolley is moving downwards along the slope right you have to calculate the time period of oscillation of this pendulum right so for that we have to calculate the the the forces actually on the pendulum so the weight m jaggs downwards or let me not write down the forces let me draw the only the acceleration which are acting on this pendulum so we have got g acting downwards and we have got two components to this theta the component which is perpendicular to this incline which is g cross theta and along the incline will be g sign theta now this is the acceleration a with which this pendulum is subjected to right so we have got two accelerations that is being acted on this pendulum which is this a and this g right so we can say that the resultant acceleration a effective we can say that is acting here will be equal to the vector addition of this two right so you can take modulus of maybe g minus a right now this is equal to root over of g squared thus here a will be equal to g sine theta right theta whole square minus to g g sine theta times cause of the angle between them right cause of 90 degree minus theta so this is this angle right that is considered and this square root will go completely right so so if you evaluate, you will have this equal to g if you take common, g squared, let's say, 1 plus sine square theta minus, this is also equal to sine theta, right? 2, sine squared theta, right? and this gives you, if we take outside, then this will be equal to one minus sine square theta, which is equal to g cost theta, right? this is what is asking the question, and if you see this is given in option c itself...

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