Question

A man of mass \(60 \mathrm{~kg}\) standing on a platform executing S.H.M. in the vertical plane. The displacement from the mean position varies as \(y=0.5 \sin (2 \pi f t)\). The value of \(f\), for which the man will feel weightlessness at the highest point is : \((y\) is in metres) (1) \(\frac{\mathrm{g}}{4 \pi}\) (2) \(4 \pi \mathrm{g}\) (3) \(\frac{\sqrt{2 g}}{2 \pi}\) (4) \(2 \pi \sqrt{2 g}\)

Short answer

The value of frequency \(f \) for which the man will feel weightlessness is \( \frac{\sqrt{2g}}{2 \pi} \).

Step-by-step solution

Understand the Condition for Weightlessness

Weightlessness occurs when the normal force acting on the man becomes zero. This happens at the highest point of the Simple Harmonic Motion (S.H.M.).

Determine the Maximum Acceleration

In S.H.M., the acceleration at the highest point is given by the formula: \[ a_{max} = \frac{d^2y}{dt^2} = - (2 \pi f)^2 y_{max} \] Since the displacement equation is given by \[ y = 0.5 \sin(2 \pi f t), \] we have \[ y_{max} = 0.5 \text{ meters}. \]

Relate Maximum Acceleration to Weightlessness

For the man to feel weightless, the maximum acceleration must be equal to gravitational acceleration (\[ g \]). Thus, \[ (2 \pi f)^2 y_{max} = g. \]

Solve for Frequency f

Plug in the values into the equation: \[ (2 \pi f)^2 (0.5) = g. \] Rearranging the equation to solve for \[ f, \] we get\[ f = \frac{1}{2 \pi} \sqrt{\frac{g}{0.5}}. \]

Simplify the Expression

Simplify the expression to find the specific value for \[ f: \] \[ f = \frac{\sqrt{2g}}{2 \pi}. \]

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of oscillatory motion where an object moves back and forth around an equilibrium position.
In SHM, the restoring force acting on the object is directly proportional to its displacement but in the opposite direction.
This force pulls the object back to the mean position. The displacement in SHM can be expressed using periodic functions like sine or cosine.

Weightlessness

Weightlessness is a sensation experienced when the normal force exerted by the ground (or a surface) on a body is zero.
This typically happens during free fall or when the only force acting on the body is gravity. In our SHM problem, a man feels weightless at the highest point because his acceleration equals gravitational acceleration, making the ground's support force zero.
This unique condition allows us to relate his maximum SHM acceleration to gravity.

Frequency in SHM

Frequency in SHM refers to the number of oscillations per unit time, typically measured in Hertz (Hz).
In the given problem, the displacement equation is provided as a sine function dependent on frequency, which allows us to derive other parameters like maximum displacement and acceleration.
To solve the problem, we set the maximum SHM acceleration equal to gravitational acceleration, allowing us to determine the specific frequency at which weightlessness is achieved.

Gravitational Acceleration

Gravitational acceleration, denoted as \( g \), is the acceleration due to Earth's gravity. Its approximate value is 9.8 m/s².
In our exercise, this value is crucial as we equate it to the maximum SHM acceleration to find the frequency. The man feels weightless when his acceleration due to SHM equals \( g \).
This relationship helps us solve for the frequency using the provided displacement equation and known physical constants.