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प्रश्न
Answer the following questions:
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
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.
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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संबंधित प्रश्न
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
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- Force acting is directly proportional to displacement from the mean position and opposite to it.
- Motion is periodic.
- Acceleration of the oscillator is constant.
- The velocity is periodic.
Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.
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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:
