Advertisements
Advertisements
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
Advertisements
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
APPEARS IN
संबंधित प्रश्न
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.
The phase difference between displacement and acceleration of a particle performing S.H.M. is _______.
(A) `pi/2rad`
(B) π rad
(C) 2π rad
(D)`(3pi)/2rad`
The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)
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?
The cylindrical piece of the cork of density of base area A and height h floats in a liquid of density `rho_1`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
`T = 2pi sqrt((hrho)/(rho_1g)`
where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).
A clock regulated by seconds pendulum, keeps correct time. During summer, length of pendulum increases to 1.005 m. How much will the clock gain or loose in one day?
(g = 9.8 m/s2 and π = 3.142)
Define practical simple pendulum
If the particle starts its motion from mean position, the phase difference between displacement and acceleration is ______.
A simple pendulum has a time period of T1 when on the earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of `"T"_2 // "T"_1` is ______.
If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be ______.
The period of oscillation of a simple pendulum of constant length at the surface of the earth is T. Its time period inside mine will be ______.
A particle executing S.H.M. has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s2. The period of oscillation is ______.
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.
Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force
The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?
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 ______.
