Q. 16 A4.0( 7 Votes )

Answer the follow

Answer :

In case of a simple pendulum the restoring force which is causing simple harmonic motion of bob of pendulum is


F = -mgSin𝜽


Where m is the mass of the Bob, g is acceleration due to gravity and 𝜽 is the angle made by string of pendulum with the vertical


As shown in the figure



We know force on a Body is


F = ma


Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as


a = F/m


so acceleration of bob is


a = -mgSin𝜽/m = -gSin𝜽


(negative sign because acceleration is in direction opposite to displacement)


when 𝜽 is very small we approximate value of sin𝜽 to 𝜽, for small oscillations


sin𝜽 𝜽


acceleration of bob is


a = -g𝜽


further we know displacement can be expressed as


x = l𝜽


where x is the displacement of bob making an circular arc and 𝜽 is the angle covered and l is the length of pendulum so we get


𝜽 = x/l


And acceleration of pendulum as


a = -(g/l)x


in case of a mass attached to a spring


as shown in figure



suppose extension in spring is x so restoring force acting on mass will be


F = -Kx


We know force on a Body is


F = ma


Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as


a = F/m


so acceleration of mass is


a = -Kx/m


or a = -(K/m)x


now in both the cases acceleration is proportional to mass since in case of pendulum acceleration due to gravity g, length of pendulum is constant so acceleration of bob of pendulum is proportional to displacement from mean position and in case of spring, the spring constant K and mass m are constant so acceleration of mass is proportional to displacement of mass from mean position


so both are undergoing simple harmonic motion and we have relation between acceleration and displacement as


a = -ω2x


where a is the acceleration of Body undergoing Simple Harmonic Motion with angular frequency ω, x is displacement from equilibrium or mean position


so for pendulum we have


ω2 = g/l


or angular frequency of oscillation as



We know relation between time period T and angular frequency ω T = 2π/ω


So we have the time period of the oscillation as



Now in case of mass attached to spring


ω2 = K/m


or angular frequency of oscillation as



We know relation between time period T and angular frequency ω T = 2π/ω


So we have the time period of the oscillation as



As we can see in case of pendulum term of mass gets cancelled out but not so in case of a mass attached to a spring


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A tunnel is dug tPhysics - Exemplar

The length of a sPhysics - Exemplar

One end of a V-tuPhysics - Exemplar

Draw a graph to sPhysics - Exemplar

A simple pendulumPhysics - Exemplar

A cylindrical logPhysics - Exemplar

A body of mass m Physics - Exemplar

Find the displacePhysics - Exemplar

Consider a pair oPhysics - Exemplar

A mass of 2 kg isPhysics - Exemplar