Advertisements
Advertisements
Question
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
Advertisements
Solution 1
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.
Solution 2
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
APPEARS IN
RELATED QUESTIONS
If the metal bob of a simple pendulum is replaced by a wooden bob of the same size, then its time period will.....................
- increase
- remain same
- decrease
- first increase and then decrease.
When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum.
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.
Answer the following questions:
What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = acos (ωt+θ) and note that the initial velocity is negative.]
Define practical simple pendulum
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?
Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.
If the particle starts its motion from mean position, the phase difference between displacement and acceleration is ______.
If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be ______.
The relation between acceleration and displacement of four particles are given below: Which one of the particles is executing simple harmonic motion?
Which of the following statements is/are true for a simple harmonic oscillator?
- 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.
A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) when U0 and α are constants. Find the time period of small oscillations.
A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. `T = 2πsqrt(m/(Apg))` where m is mass of the body and ρ is density of the liquid.
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 ______.
