Take a metre scale and measure the length of the string from the point of suspension to the lower tip of the bob \(\left(\ell_{1}\right)\) (b). Now, place the bob over a meter scale and hold it in position with two wooden blocks or stiff cardboards and measure the diameter (D) of the bob (c). Calculate the radius \(\mathrm{R}\) of the bob by dividing diameter by 2 (a). Then the length of the pendulum \(\ell=\left(\ell_{1}-\mathrm{R}\right)(\mathrm{d})\). Consider the formula \(\mathrm{T}\) \(=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}\) and find the time period of the simple pendulum by substituting the value of \({ }^{\prime} \ell^{\prime}(\mathrm{e})\).

Short Answer

Expert verified
2. How can you measure the diameter of the pendulum bob? 3. What formula is used to calculate the radius of the pendulum bob? 4. How is the length of the pendulum calculated? 5. Which formula is used to find the time period of a simple pendulum?

Step by step solution

01

1. Measure the length of the string

Measure the length of the string from the point of suspension to the lower tip of the bob (\(\ell_1\)) using a metre scale.
02

2. Measure the diameter of the bob

Place the bob over a meter scale and hold it in position with two wooden blocks or stiff cardboards. Measure the diameter (D) of the bob.
03

3. Calculate the radius of the bob

Calculate the radius (R) of the bob by dividing the diameter by 2: \[R = \frac{D}{2}\]
04

4. Calculate the length of the pendulum

Calculate the length of the pendulum (\(\ell\)) by subtracting the radius (R) from the length of the string (\(\ell_1\)): \[\ell = \ell_1 - R\]
05

5. Calculate the time period of the simple pendulum

Use the formula for the time period of a simple pendulum to find the time period (T): \[T = 2\pi\sqrt{\frac{\ell}{g}}\] Here, \(g\) is the acceleration due to gravity (\(9.81 \, m/s^2\)). Substitute the calculated values of \(\ell\) and \(g\) into the formula to find the time period (T).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pendulum Length Measurement
Understanding how to accurately measure the length of a simple pendulum is crucial, as it directly influences the pendulum's time period. To ensure an exact measurement, use a meter scale or a ruler with a fine graduation. It's imperative to measure from the point of suspension—the fixed point where the pendulum swings—to the very bottom of the bob, effectively the center of mass. However, since the bob has a sizeable dimension, only measuring to its bottom will not suffice for precision.

Once you obtain this measurement, denoted as \(\ell_1\), you then need to account for the bob's radius to determine the actual length of the pendulum. This is done by carefully measuring the bob’s diameter with the same meticulous approach you used for the string, preferably using a caliper for higher precision, then dividing the diameter by two to obtain the radius. The critical step here is deducting this radius from your initial measurement, \(\ell_1\), which then gives you the precise pendulum length, \(\ell\). This adjusted length is the one used when calculating the time period of oscillation.
Time Period Calculation
Calculating the time period of a simple pendulum is fundamental in the study of harmonic motion. The time period, \(T\), refers to the duration it takes for the pendulum to complete one full swing back and forth. The formula to calculate this is \(T = 2\pi\sqrt{\frac{\ell}{g}}\), where \(\ell\) is the effective length of the pendulum and \(g\) represents the acceleration due to gravity, usually approximated to \(9.81 m/s^2\).

However, the precision of \(T\) hinges on the accurate measurement of \(\ell\) as mentioned earlier. Even slight deviations in measuring the length can lead to significant errors in the calculated time period. Once you have \(\ell\) nailed down, the computation of \(T\) becomes a simple mathematical exercise. It's critical to ensure that the calculations are carried out with the correct units to avoid any further errors. Understanding this relationship between the pendulum’s length and its time period is essential in experiments and practical applications of pendulums.
Radius of Bob Calculation
The radius of the bob is an important factor in the experiment of a simple pendulum, as it influences the accurate measurement of the pendulum's length. To calculate the bob's radius, which we denote as \(R\), you first need to measure the diameter of the bob. Position the bob so it is stable and horizontal, then measure across its widest part. A common tool for this would be a pair of vernier calipers for the most accurate measurement.

Once you have the diameter, the radius is simply half of this value, represented mathematically as \(R = \frac{D}{2}\). This step might seem minor, but any error in calculating the radius can lead to a cascade of inaccuracies in further calculations, particularly in the pendulum's length and subsequently, its time period. Hence, attention to detail here is indispensable for constructing reliable experimental data.

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Most popular questions from this chapter

\(\mathrm{A} \rightarrow \mathrm{c} \quad\) The motion of a shell fired from artillery gun is curvilinear. \(\mathrm{B} \rightarrow \mathrm{d} \quad\) A stone dropped from a tower moves vertically downwards. It is in rectilinear motion. \(\mathrm{C} \rightarrow \mathrm{b} \quad\) The wings of a ceiling fan undergo rotatory motion. \(\mathrm{D} \rightarrow \mathrm{g} \quad\) Acceleration, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{If}}\) body moves with uniform velocity, then \(\mathrm{v}=\mathrm{u} \Rightarrow \mathrm{a}=0 .\)

We know, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) The time period of seconds pendulum, \(\mathrm{T}_{1}=2 \mathrm{~s}\) The length of the second pendulum \(\ell_{1}=100 \mathrm{~cm}=1 \mathrm{~m}\) Now, the new length of the pendulum \(\ell_{2}=2 \ell_{1}\) \(\Rightarrow \ell_{2}=200 \mathrm{~cm}=2 \mathrm{~m}\) Let the new time period of the pendulum be \(=\mathrm{T}_{2}\) \(\Rightarrow \mathrm{T} \alpha \sqrt{l}\) \(\Rightarrow \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{l_{1}}{l_{2}}} \Rightarrow \mathrm{T}_{2}=\mathrm{T}_{1} \sqrt{\frac{l_{2}}{l_{1}}}\) \(\mathrm{T}_{2}=2 \times \sqrt{\frac{2 l_{1}}{l_{1}}}=2 \sqrt{2} \mathrm{~s}\) The time period becomes \(2 \sqrt{2}\) times the original one.

The speed and average speed of the vehicle can be equal if the vehicle moves with uniform speed or constant speed.

Consider a body moving with initial velocity u. Let its velocity change to \(\mathrm{v}\), in time ' \(\mathrm{t}^{\prime}\). Then, the change in velocity is \(=\mathrm{v}-\mathrm{u}\). The change in velocity per unit time \(=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) By definition, the change in velocity per unit time is acceleration, a. Thus, \(\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\) or \(\mathrm{v}-\mathrm{u}=\) at \(\mathrm{v}=\mathrm{u}+\mathrm{at}\)

Unit of time in SI and CGS systems is second (s). Unit of speed in SI and CGS systems are \(\mathrm{m} \mathrm{s}^{-1}\) and \(\mathrm{cm} \mathrm{s}^{-1}\) respectively.

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