Power Laws

Problem 1. The period T seconds of a simple pendulum of length L cm satisfies the equation

T = \alpha \sqrt{L}

for some positive constant α. Experimental measurements yielded the following results.

L (cm)203040506070
T (s)0.891.101.261.411.551.67

Draw a straight-line graph of T2 against L. Hence, estimate the value of α.

(Click for Solution)

Solution. We first plot the values as follows, correct to the nearest 0.025 for the T2-axis.

L203040506070
T20.8001.2001.6002.0002.4002.800

Squaring both sides, T2 = α2 · L. Hence, we graph as follows.

We estimate the gradient of the straight-line graph by

\displaystyle \alpha^2 \approx \frac{2.000}{50} = 0.04.

Taking square roots, α ≈ 0.2.

Problem 2. For a thin converging lens with focal length f cm, the object distance u cm and the image distance v cm are related by

\displaystyle \frac 1u + \frac 1v = \frac 1f.

The following values were obtained in an experiment

u (cm)202530405060
v (cm)60.037.530.024.021.420.0

Draw a straight-line graph of \displaystyle \frac 1v and \displaystyle \frac 1u. Hence, estimate the value of f correct to 3 significant figures.

(Click for Solution)

Solution. We first plot the values as follows.

1/u0.05000.04000.03350.02500.02000.0165
1/v0.01650.02650.03350.04150.04650.0500

Shifting 1/u to the right,

\displaystyle \frac 1v = -\frac 1u + \frac 1f.

Hence, we graph as follows.

We estimate the (1/v)-intercept of the straight-line graph by

\displaystyle \frac 1f \approx 0.0665 \quad \Rightarrow \quad f \approx 15.0.

Problem 3. The population P of a city is modelled by

P = A \cdot 10^{kt}

where t is the time in years after a reference year and A, k are positive constants. The following data was recorded.

t (years)5101520
P17783162562310 000

Draw a straight-line graph of lg P against t. Hence, estimate the values of A and k.

(Click for Solution)

Solution. We first plot the values as follows.

t5101520
lg P3.2503.5003.7504.000

Taking logarithms on both sides,

\lg P = kt + \lg A.

Hence, we graph as follows.

We estimate the gradient of the straight-line graph by

\displaystyle k \approx \frac{0.75}{15} = 0.05,\quad \lg A \approx 3.

Taking exponentials, A ≈ 103 = 1000.

Problem 4. In a second-order chemical reaction, the concentration x units of a reactant at time t seconds satisfies the equation

\displaystyle \frac 1x = k t + \frac 1{x_0},

where k is the rate constant and x0 is the initial concentration. The following data was recorded.

t (s)246810
x (units)0.4550.4170.3850.3570.333

Draw a straight-line graph of \displaystyle \frac 1x against t. Hence, estimate the values of k and x0.

(Click for Solution)

Solution. We first plot the values as follows.

t246810
1/x2.2002.4002.6002.8003.000

Hence, we graph as follows.

We estimate the gradient of the straight-line graph by

\displaystyle k \approx \frac{0.800}{8} = 0.1,\quad \frac 1{x_0} \approx 2.

Taking reciprocals, x0 ≈ 1/2.0 = 0.5.

—Joel Kindiak, 7 Apr 26, 1808H

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