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प्रश्न
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 T is greater than `2pisqrt(1/g)` Think of a qualitative argument to appreciate this result.
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उत्तर १
In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
F = –mg sinθ
Where,
F = Restoring force
m = Mass of the bob
g = Acceleration due to gravity
θ = Angle of displacement
For small θ, sinθ = θ
For large θ, sinθ is greater than θ.
This decreases the effective value of g.
Hence, the time period increases as:
`T = 2pi sqrt(1/g)`
Where, l is the length of the simple pendulum
उत्तर २
The restoring force for the bob of the pendulum is given by
`F = -mg sintheta`
if `theta` is small thensin `theta = theta = y/l` `:. F = -(mg)/l y`
i.e the motion is simple harmonic and time period is` T = 2pi sqrt(1/g)`
Clearly, the above formula is obtained only if we apply the approximation `sin theta ~~ theta`
For large angles this approximation is not valid and T is greater than `2pi sqrt(1/g)`
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 ______.
When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum.
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.
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
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In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
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- Motion is periodic.
- Acceleration of the oscillator is constant.
- The velocity is periodic.
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A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Figure). The amplitude is θ0. The string snaps at θ = θ0/2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θo to be small so that sin θo = θo and cos θo = 1.
A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then ______.
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is `x/2` times its original time period. Then the value of x is ______.
