Advertisements
Advertisements
प्रश्न
If the inertial mass and gravitational mass of the simple pendulum of length l are not equal, then the time period of the simple pendulum is
विकल्प
T = `2π sqrt(("m"_"i""l")/("m"_"g""g"))`
T = `2π sqrt(("m"_"g""l")/("m"_"i""g"))`
T = `2π "m"_"g"/"m"_"i" sqrt("l"/"g")`
T = `2π "m"_"i"/"m"_"g" sqrt("l"/"g")`
Advertisements
उत्तर
T = `2π sqrt(("m"_"i""l")/("m"_"g""g"))`
APPEARS IN
संबंधित प्रश्न
The average displacement over a period of S.H.M. is ______.
(A = amplitude of S.H.M.)
A particle executes simple harmonic motion Let P be a point near the mean position and Q be a point near an extreme. The speed of the particle at P is larger than the speed at Q. Still the particle crosses Pand Q equal number of times in a given time interval. Does it make you unhappy?
A student says that he had applied a force \[F = - k\sqrt{x}\] on a particle and the particle moved in simple harmonic motion. He refuses to tell whether k is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle.
The time period of a particle in simple harmonic motion is equal to the time between consecutive appearances of the particle at a particular point in its motion. This point is
The displacement of a particle in simple harmonic motion in one time period is
The displacement of a particle is given by \[\overrightarrow{r} = A\left( \overrightarrow{i} \cos\omega t + \overrightarrow{j} \sin\omega t \right) .\] The motion of the particle is
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run
A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t = 4 s.
Define the time period of simple harmonic motion.
What is the ratio of maxmimum acceleration to the maximum velocity of a simple harmonic oscillator?
