(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly

(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly

**(a) 35.2 cm (b) 23 cm (c) 15.3 cm**

**(d)12.4 cm (e) 9.4 cm **

The angular frequency (ω) of oscillation of the pendulum is given by

ω = 2π/T = 2π/2 = π rad/s.

The bob has maximum velocity (v_{max}) in the mean position and is given by

v_{max }=_{ } ωA where A is the amplitude.

Therefore, v_{max} = π × 0.1 = 0.1 π ms^{–1}** **

On getting detached from the string, the bob moves like a projectile shot horizontally from a height of 0.45 m with a velocity of 0.1 π ms^{–1}.** **Its time of flight** **(t)** **is obtained from the vertical displacement of 0.45 m:

0.45 = 0 + ½ gt^{2}. (Note that the initial vertical velocity is zero and the vertical acceleration is g, which we may take as 10 ms^{–2}). This gives t^{2} =^{ }0.09 so that t = 0.3 s.

The horizontal distance covered by the bob during this time is 0.1 π × 0.3 = 0.094 m = 9.4 cm.

[Note that the mass of the bob does not come into the picture and it just serves as a distraction].

Now, consider the following MCQ :

**A large horizontal surface moves up and down simple harmonically with an amplitude of 1 cm. If a mass of 3 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of the SHM will be**

**(a) 5 Hz (b) 2 Hz (c) 8 Hz (d) 10Hz (e) 15 Hz**

The mass will remain in contact with the surface if the maximum acceleration produced in the simple harmonic motion does not exceed the acceleration due to gravity (g).

Therefore, we have

ω^{2}A = g where ω is the angular frequency of the SHM.

From this ω = √(g/A) = √(10/ 0.01) = 10×√10.

The maximum frequency (linear) of oscillations is therefore given by

n = ω/2π = (10×√10)/2π = 5 Hz.** **

[Note that the mass of the body placed on the surface does not come into the picture and it just serves as a distraction].

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