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A Time Period of a Particle in Shm Depends on the Force Constant K And Mass M Of the Particle: `T = 2pi Sqrt(M/K)` A Simple Pendulum Executes Shm Approximately. Why Then is the Time

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प्रश्न

Answer the following questions:

A time period of a particle in SHM depends on the force constant and mass of the particle: `T = 2pi sqrt(m/k)` A simple pendulum executes SHM approximately. Why then is the time 

 

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उत्तर १

The time period of a simple pendulum, `T = 2pi sqrt(m/k)`

For a simple pendulum, k is expressed in terms of mass m, as: `k prop m` 

`m/k =` Constant

Hence, the time period T, of a simple pendulum is independent of the mass of the bob.

 

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उत्तर २

In case of a spring, k does not depend upon m. However, in case of a simple pendulum, k is directly proportional to m and hence the ratio m/k is constant quantity

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Oscillations - Exercises [पृष्ठ ३६०]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 11
पाठ 13 Oscillations
Exercises | Q 16.1 | पृष्ठ ३६०

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संबंधित प्रश्‍न

The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is ______.


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The phase difference between displacement and acceleration of a particle performing S.H.M. is _______.

(A) `pi/2rad`

(B) π rad

(C) 2π rad

(D)`(3pi)/2rad`


A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.


The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (on the surface of earth is 9.8 ms–2)


Answer the following questions:

The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that is greater than `2pisqrt(1/g)`  Think of a qualitative argument to appreciate this result.


Answer the following questions:

What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?


The cylindrical piece of the cork of density of base area and height floats in a liquid of density `rho_1`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

`T = 2pi sqrt((hrho)/(rho_1g)` 

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).


Show that motion of bob of the pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends?


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  2. Motion is periodic.
  3. Acceleration of the oscillator is constant.
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When will the motion of a simple pendulum be simple harmonic?


The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?


A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.


In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:


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(In the frame of the trolley)

 


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