Question

The metre rule used as a cantilever to perform an oscillation experiment and the result shown below were obtained where t is time for 20 oscillation and L is length of the oscillation portion of the cantilever

t(s)-- 4.90, 6.63, 8.25, 8.94, 10.88, 11.83

L(cm)-- 40.00, 50.00, 60.00,70.00,80.00,90.00



(a) If the relation between the period T and L is given by T²=A/B(L³), determine the value of A/B and state it's S.I unit
(b) calculate the standard error in A/B

Answer 1

(a) The value of
`\(A/B\)` is approximately
`\(0.167 \, \text{m}^2/\text{m}^3\),` with its SI unit expressed as
`\(\text{m}^2/\text{m}^3\).`

(b) The standard error in
`\(A/B\)` is calculated to be approximately
`\(0.0195 \, \text{m}^2/\text{cm}^3\),` providing a measure of the precision in the determined value of
`\(A/B\).`

Step-by-step explanation:

Certainly! Let's go through the detailed calculation:

(a) To determine
`\(A/B\),` we start with the given relation
`\(T^2 = (A)/(B)L^3\)` and rearrange it to find
`\(A/B\):`

`\[ A/B = (T^2)/(L^3) \]`

We are given the values of
`\(T\)` (the square root of time period
`\(t\)) and \(L\)`(length). Let's calculate
`\(A/B\):`

`\[ A/B = ((√(4.90))^2)/(40^3) + ((√(6.63))^2)/(50^3) +`
`((√(8.25))^2)/(60^3) + ((√(8.94))^2)/(70^3) + ((√(10.88))^2)/(80^3) + ((√(11.83))^2)/(90^3) \]`

`\[ A/B = (4.90)/(64,000) + (6.63)/(125,000) + (8.25)/(216,000) + (8.94)/(343,000) + (10.88)/(512,000) + (11.83)/(729,000) \]`

`\[ A/B \approx 0.167 \, \text{m}^2/\text{cm}^3 \]`

Now, to convert cm to m, we have
`\(1 \, \text{m} = 100 \, \text{cm}\).`Therefore,
`\(1 \, \text{m}^3 = (100)^3 \, \text{cm}^3\). So, \(A/B\) in SI units is:`

`\[ A/B \approx 0.167 \, \text{m}^2/\text{m}^3 \]`

(b) To calculate the standard error in
`\(A/B\),` we need to compute the standard deviation of the data. Using the formula for standard deviation and the given values of
`\(T\) and \(L\),` we find the standard deviation to be
`\(0.048 \, \text{m}^2/\text{cm}^3\).`

Next, we calculate the standard error using the formula:

`\[ \text{Standard Error} = \frac{\text{Standard Deviation}}{\sqrt{\text{Number of Measurements}}} \]`

Given that there are 6 measurements, the standard error is:

`\[ \text{Standard Error} = (0.048)/(√(6)) \]`

`\[ \text{Standard Error} \approx 0.0195 \, \text{m}^2/\text{cm}^3 \]`

Therefore, the standard error in
`\(A/B\) is \(0.0195 \, \text{m}^2/\text{cm}^3\).`