Question

Figure 15-25shows plots of the kinetic energy K versus position x for three harmonic oscillators that have the same mass. Rank the plots according to (a) the corresponding spring constant and (b) the corresponding period of the oscillator, greatest first.

Short answer

a) Ranking the plots to corresponding spring constant is$k_A>k_B>k_C$ .

b) Ranking the plots corresponding period of oscillator is$T_C>T_B>T_A$ .

Step-by-step solution

The given data 

The graph of K versus x for three harmonic oscillators is given.

Understanding the concept of energy and period in SHM

We can use the law of conservation of energy and write for the maximum kinetic energy of an object in terms of the spring constant. From the equation and graph given, we can compare the proportionality of K and k

and can rank the plots according to spring constants. Using the equation of period and using the relation between T and k we can rank the plots for T.

Formulae:

The kinetic energy of a spring-system,$K=\frac{1}{2}k{x}_m^2$ (i)

The period of an oscillation in SHM, $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$ (ii)

Calculation of the ranking the plots to corresponding spring constant

a)

From equation (i), we can see that K is directly proportional to spring constant k.

$K\alpha k$

Therefore, the ranking of the plots according to the spring constant is$k_A>k_B>k_C$ .

Calculation of the ranking the plots corresponding period of oscillator

b)

From equation (ii), the period T is found to be inversely proportional to$\sqrt{k}$, that is given as:

$T\alpha \frac{1}{\sqrt{k}}$

Since the ranking of plots according to spring constant is role="math" localid="1657261768550" $k_A>k_B>k_C$

Hence, the ranking of plots according to the period of the oscillator is$T_C>T_B>T_A$ .