91Ó°ÊÓ

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

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,$\mathrm{Ó\neg }=\sqrt{\frac{k}{m}}$ (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)

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 ${\mathrm{Ó\neg }}_1={\mathrm{Ó\neg }}_2={\mathrm{Ó\neg }}_3$.

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 $U_3>U_2=U_1$.

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 $K.E_1>K.E_2>K.E_3$.

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 $v_1>v_2>v_3$.

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 $K.E_{max1}>K.E_{max3}>K.E_{max2}$.