Question

Figure 15-24shows the x(t) curves for three experiments involving a particular spring–box system oscillating in SHM. Rank the curves according to (a) the system’s angular frequency, (b) the spring’s potential energy at time t=0, (c) the box’s kinetic energy att=0, (d) the box’s speed att=0, and (e) the box’s maximum kinetic energy, greatest first.

Short answer

a) Ranking of curves according to system’s angular frequency is${\omega }_1={\omega }_2={\omega }_3$

b) Ranking of curves according to springs potential energy at t = 0 is $U_3>U_2=U_1$.

c) Ranking of curves according to box’s kinetic energy at t = 0 is $K.E_1>K.E_2>K.E_3$.

d) Ranking of curves according to box’s speed at t = 0 is $v_1>v_2>v_3$.

e) Ranking of curves according to box’s maximum kinetic energy is role="math" localid="1657260771388" $K.E_{max1}>K.E_{max3}>K.E_{max2}$.

Step-by-step solution

The given data 

The graph of position versus time for SHM of spring-box system is given.

Understanding the concept of SHM of a particle

Using the formula for angular frequency which is related to spring constant and mass, we can rank the curves. From the spring’s potential energy formula, we can rank the curves, and from the formula of kinetic energy, we can rank the curves for kinetic energy. We can also rank the speed of the box. For ranking maximum kinetic energy, we consider the amplitude of the curves.

Formulae:

The angular frequency of a body in SHM,$\omega =\sqrt{\frac{k}{m}}$ (i)

The potential energy of a spring system,$U=\left(\frac{1}{2}\right)k{x}^2$ (ii)

The kinetic energy of a body in SHM, $K.E=\left(\frac{1}{2}\right)m{v}^2$ (iii)

Calculation of the ranking of curves according to the system’s angular frequency

a)

As we know the mass of the box and the spring constant is the same for the same oscillatory system while doing the experiment, therefore, the angular frequency will be the same for these curves considering equation (i).

Hence, ranking of angular frequencies is ${\omega }_1={\omega }_2={\omega }_3$.

Calculation of the ranking of curves according to potential energy of the spring

b)

From equation (ii), we can see that the potential energy of the spring depends on the displacement.

Again, in the graph at t = 0 we get the displacement of curve 3 is greater than the other two. The displacements of curve 1 and 2 are same.

Hence, the ranking of potential energies is $U_3>U_2=U_1$.

Calculation of the ranking of curves according to box’s kinetic energy

c)

Slope of the curves gives the velocity, $v=\frac{∆x}{∆t}$

From the graph, the rank of the velocities at t = 0 can be given as:$v_1>v_2>v_3$.

Therefore, the ranking of kinetic energies using equation (iii) is $K.E_1>K.E_2>K.E_3$.

Calculation of the ranking of curves according to box’s speed

d)

From part (c), we have seen that ranking of velocities as:$v_1>v_2>v_3$ , and since all have same direction the ranking of speed is same as the velocities at t = 0.

Therefore, the ranking of speed is $v_1>v_2>v_3$.

Calculation of the ranking of curves according to maximum kinetic energy of the box

e)

The kinetic energy is proportional to the amplitude of the curve, so from the graph we get,

The amplitude of the curves as:1>3>2 .

The ranking of maximum kinetic energies is $K.E_{max1}>K.E_{max3}>K.E_{max2}$.