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Chapter 3: Problem 44

A simple pendulum that has time period of \(2 \mathrm{~s}\) is called seconds pendulum. Its length is approximately \(100 \mathrm{~cm}\) or \(1 \mathrm{~m} .\)

Short Answer

Expert verified
Answer: The approximate length of a seconds pendulum, which has a time period of 2 seconds, is 100 centimeters or 1 meter.

Step by step solution

01

Understand the formula for the time period of a simple pendulum

The time period \(T\) of a simple pendulum can be determined using the formula: \(T = 2\pi\sqrt{\frac{l}{g}}\) where \(T\) is the time period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity (approximately \(9.81\mathrm{~m/s^2}\)).
02

Set up the equation for the seconds pendulum

Given that the time period of a seconds pendulum is 2 seconds, we can replace \(T\) with 2 in the formula: \(2 = 2\pi\sqrt{\frac{l}{9.81\mathrm{~m/s^2}}}\)
03

Solve the equation for the length \(l\)

Now, we need to solve the equation for \(l\). Divide both sides of the equation by \(2\pi\): \(\frac{2}{2\pi} = \sqrt{\frac{l}{9.81\mathrm{~m/s^2}}}\) Square both sides of the equation to remove the square root: \(\left(\dfrac{2}{2\pi}\right)^2 = \frac{l}{9.81}\)
04

Complete the calculations to find the length

Multiply both sides of the equation by \(9.81\mathrm{~m/s^2}\) to find \(l\): \(l = 9.81\mathrm{~m/s^2}\times\left(\dfrac{2}{2\pi}\right)^2\) \(l\approx100\mathrm{~cm}\) or \(1\mathrm{~m}\)
05

Conclusion

The length of a seconds pendulum, which has a time period of 2 seconds, is approximately 100 centimeters or 1 meter.

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Most popular questions from this chapter

The different kinds of motion are: (i) Translatory motion: \(\mathrm{A}\) bus moving on a road, the motion of a rising balloon, the free fall of a stone under gravity, the motion of a cricket ball when it is hit by a batsman are examples of translatory motion. Translatory motion is further classified as rectilinear motion and curvilinear motion. When an object moves along a straight path, its motion is said to be rectilinear motion. The marching of soldiers on a straight road, the motion of a car on a straight road, the motion of carrom board coin are examples of rectilinear motion. When an object moves along curved path, its motion is called curvilinear motion. A bus moving on a fly-over bridge, a car taking a turn, a football kicked from the ground into air all have curvilinear motion. (ii) Rotatory motion: In this type of motion, the object rotates about a fixed axis. The motion of blades of a ceiling fan, the spin motion of a top, the motion of turbine, the motion of the earth around the sun are all circular or rotatory motion. In some cases, the rotatory and translatory motions take place simultaneously. When a bicycle moves, its wheels undergo translatory and rotatory motion. (iii) Oscillatory motion: A boy on a swing moves to-and-fro (back and forth). The motion such as above, where an object moves to-and-fro is called oscillatory motion. Other examples of oscillatory motion are the motion of the pendulum of a clock, the motion of a needle of a sewing machine, the motion of a piston of an engine etc.

An object moves to and fro produces oscillatory motion. If the object moves to and fro at faster rate, it is called vibratory motion.

(a) Let the distance travelled be 'd'. The speed of \(^{\prime} A^{\prime}, v_{A}=\frac{d}{20} m s^{-1}\) The speed of ' \(\mathrm{B}^{\prime}, \mathrm{v}_{\mathrm{B}}=\frac{\mathrm{d}}{22} \mathrm{~m} \mathrm{~s}^{-1}\). \(\Rightarrow \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}}{\frac{\mathrm{d}}{22}}=\frac{22}{20}=\frac{11}{10}=11: 10\) (b) Let them run for 't's. Then, \(\mathrm{d}_{\mathrm{A}}=\mathrm{v}_{\mathrm{A}} \times \mathrm{t}=\frac{\mathrm{d}}{20} \times \mathrm{t}\) \(\mathrm{d}_{\mathrm{B}}=\mathrm{v}_{\mathrm{B}} \times \mathrm{t}=\frac{\mathrm{d}}{22} \times \mathrm{t}\) \(\Rightarrow \frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}=\frac{\frac{\mathrm{d}}{20}(\mathrm{t})}{\frac{\mathrm{d}}{22}(\mathrm{t})}=\frac{22}{20}=\frac{11}{10}=11: 10\)

Distance travelled in the first \(20 \mathrm{~min}=\) speed \(\times\) time \(=60 \times \frac{20}{60}=20 \mathrm{~km}\)

Fill in the Blanks. \(5 \mathrm{~m} \mathrm{~s}^{-2}\) \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{20-10}{2}=5 \mathrm{~m} \mathrm{~s}^{-2}\)
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