Advertisements
Advertisements
Question
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.
Advertisements
Solution 1
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
Solution 2
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)`
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.
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.
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
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 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
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?
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 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 ______.
The relation between acceleration and displacement of four particles are given below: Which one of the particles is executing simple harmonic motion?
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 ______.
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.

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 ______.
