if T = 2π√l/g and L = 100±0.1cm(limit standard error), T=2.01±0.01s(limit standard error). Calculate the value of getting and it's standard error
Well, well, well, let's crunch some numbers, shall we? According to the equation T = 2π√(L/g), where L is the length of the pendulum, and g is the acceleration due to gravity, you have values for T and L along with their standard errors. Alrighty then, let's calculate shall we?
Given:
L = (100 ± 0.1) cm
T = (2.01 ± 0.01) s
The value of g can be determined by rearranging the equation as follows:
g = (4π²L)/T²
Now, let's plug in the values:
L = (100 ± 0.1) cm
T = (2.01 ± 0.01) s
Calculating g:
g = (4π² * (100 ± 0.1)) / (2.01 ± 0.01)²
Alright, now it's time to whip out our trusty calculator and perform the calculations:
g = (4π² * 100) / 2.01²
g ≈ 9.831 m/s²
Now, for the standard error. When you multiply or divide quantities and the standard errors are given by Δa and Δb, the standard error for the result is given by the formula:
Δf/f = √((Δa/a)² + (Δb/b)²)
Alright, let's calculate for the standard error of g:
Δg/g = √((ΔL/L)² + (2ΔT/T)²)
Δg/g = √((0.1/100)² + (2 * 0.01/2.01)²)
Aw, come on, let me grab my handy-dandy calculator once again:
Δg/g ≈ √((0.001)² + (0.02/2.01)²)
Δg/g ≈ √(0.000001 + 0.000198)^2
Δg/g ≈ √0.000199^2
Δg/g ≈ 0.0141
Okay, so the value of g is approximately 9.831 m/s² and its standard error is approximately ±0.0141. Phew! That was quite the mathematical circus, wasn't it? Hope I didn't juggle too many numbers there.
To calculate the value of "g" and its standard error, we can rearrange the equation T = 2π√(L/g) to solve for "g".
Step 1: Squaring both sides of the equation yields:
T^2 = (2π√(L/g))^2
T^2 = 4π^2(L/g)
T^2 = (4π^2L)/g
Step 2: Isolating "g" on one side of the equation, we have:
g = (4π^2L)/T^2
Given:
L = 100 ± 0.1 cm (limit standard error)
T = 2.01 ± 0.01 sec (limit standard error)
Step 3: Calculate the value of "g":
Substituting the given values into the equation gives:
g = (4π^2(100))/2.01^2
g ≈ 4π^2(100)/(2.01^2)
g ≈ (4π^2(100))/(4.0401)
g ≈ (4π^2(100))/4.0401
g ≈ 100π^2/4.0401
g ≈ 78.223 cm/sec^2
Hence, the estimated value of "g" is approximately 78.223 cm/sec^2.
Step 4: Calculate the standard error of "g":
To calculate the standard error, we can use the formula:
Standard Error = (ΔL/L) * g
Using the given limit standard error ΔL/L = 0.1/100 = 0.001 and the estimated value of "g" as 78.223 cm/sec^2, we can calculate:
Standard Error = (0.001) * 78.223
Standard Error ≈ 0.078 cm/sec^2
Therefore, the standard error of "g" is approximately 0.078 cm/sec^2.
To calculate the value of "g" and its standard error using the given values of T and L, you can rearrange the equation T = 2π√(L/g) to solve for "g".
1. Start by isolating the g term in the equation:
T = 2π√(L/g)
Square both sides of the equation:
T^2 = (2π)^2(L/g)
Rearrange the equation to solve for g:
g = (4π^2L) / T^2
2. Substitute the given values into the equation:
L = 100 cm (convert to meters by dividing by 100: 100/100 = 1 m)
T = 2.01 s
g = (4π^2 * 1) / (2.01^2) [Substituting L and T into the equation]
3. Calculate the value of "g":
g = (4π^2 * 1) / (2.01^2)
≈ 9.86 m/s^2
The value of "g" is approximately 9.86 m/s^2.
To calculate the standard error, we need to consider the uncertainties in "L" and "T". Since the standard errors are given as ±0.1 cm and ±0.01 s respectively, we can use the following formula:
Standard Error (SE) = √[(∂g/∂L)^2 * (SE_L)^2 + (∂g/∂T)^2 * (SE_T)^2]
Where:
∂g/∂L = Partial derivative of "g" with respect to "L"
∂g/∂T = Partial derivative of "g" with respect to "T"
SE_L = Standard Error of "L"
SE_T = Standard Error of "T"
Taking the partial derivatives:
∂g/∂L = (4π^2) / T^2
∂g/∂T = (-2 * 4π^2 * L) / T^3
Substituting the given standard errors:
SE_L = ±0.1 cm (convert to meters by dividing by 100: 0.1/100 = 0.001 m)
SE_T = ±0.01 s
SE = √[(∂g/∂L)^2 * (SE_L)^2 + (∂g/∂T)^2 * (SE_T)^2]
SE = √[((4π^2) / T^2)^2 * (0.001)^2 + ((-2 * 4π^2 * L) / T^3)^2 * (0.01)^2]
Calculating the standard error using the given values:
SE ≈ 0.032 m/s^2
The standard error of "g" is approximately 0.032 m/s^2.