Advertisements
Advertisements
प्रश्न
State the laws of the simple pendulum?
Advertisements
उत्तर
- Law of length: For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of the length of the pendulum.
`"T" ∝ sqrt"l"` - Law of acceleration: For a fixed length, the time period of a simple pendulum is inversely proportional to the square root of acceleration due to gravity.
`"T" ∝ 1/sqrt"g"` - The period of a simple pendulum is independent of its mass.
- A simple pendulum's period is independent of its amplitude (for small amplitudes).
APPEARS IN
संबंधित प्रश्न
A particle executes simple harmonic motion. If you are told that its velocity at this instant is zero, can you say what is its displacement? If you are told that its velocity at this instant is maximum, can you say what is its displacement?
In a simple harmonic motion
(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy
(c) the minimum potential energy equals the maximum kinetic energy
(d) the maximum potential energy equals the minimum kinetic energy
A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a second pendulum at a place where g = π2 m/s2.
A simple pendulum is constructed by hanging a heavy ball by a 5.0 m long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is 1.67 m/s2?
A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with and acceleration a0(b) is going down with an acceleration a0 and (c) is moving with a uniform velocity.
A particle is subjected to two simple harmonic motions, one along the X-axis and the other on a line making an angle of 45° with the X-axis. The two motions are given by x = x0 sin ωt and s = s0 sin ωt. Find the amplitude of the resultant motion.
A simple pendulum has a time period T1. When its point of suspension is moved vertically upwards according to as y = kt2, where y is the vertical distance covered and k = 1 ms−2, its time period becomes T2. Then, T `"T"_1^2/"T"_2^2` is (g = 10 ms−2)
What is an epoch?
Consider the Earth as a homogeneous sphere of radius R and a straight hole is bored in it through its centre. Show that a particle dropped into the hole will execute a simple harmonic motion such that its time period is
T = `2π sqrt("R"/"g")`
Displacement vs. time curve for a particle executing S.H.M. is shown in figure. Choose the correct statements.
- Phase of the oscillator is same at t = 0 s and t = 2s.
- Phase of the oscillator is same at t = 2 s and t = 6s.
- Phase of the oscillator is same at t = 1 s and t = 7s.
- Phase of the oscillator is same at t = 1 s and t = 5s.
