Question

In fig.15-28, a spring–block system is put into SHM in two experiments. In the first, the block is pulled from the equilibrium position through a displacement and then released. In the second, it is pulled from the equilibrium position through a greater displacement${\mathit{d}}_{\mathbf{2}}$ and then released. Are the (a) amplitude, (b) period, (c) frequency, (d) maximum kinetic energy, and (e) maximum potential energy in the second experiment greater than, less than, or the same as those in the first experiment?

Short answer

a) The amplitude of second experiment is greater than the first experiment.

b) The period for both the experiments is same.

c) The frequency for both the experiments is same.

d) The maximum kinetic energy of second experiment is greater than the first experiment.

e) The maximum potential energy of second experiment is greater than the first experiment

Step-by-step solution

The given data 

The figure for the block-spring system is given.

Understanding the concept of equations of a SHM

We can use the formulae for amplitude, period, frequency, kinetic energy, and potential energy related to displacement. Comparing the displacements of the system in both experiments, we can compare the above quantities.

Formulae:

The displacement equation of a body in SHM,$x=x_m\cos \left(\omega t+\varphi \right)$ (i)

The acceleration of the body in SHM, $SHM,a=-{\omega }^{2}x$ (ii)

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

The frequency of the oscillation in SHM, $\mathrm{SHM},\mathrm{f}=\frac{1}{\mathrm{T}}$ (iv)

The maximum kinetic energy of a system, $\mathrm{K}.{\mathrm{E}}_{\max }=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2$ (v)

The maximum potential energy of a system, ${\mathrm{U}}_{\max }=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2$ (vi)

Calculation of the amplitudes of the experiments

a)

Here, displacements are $d_1$and$d_2$ Now, from equation (i), we can get that

$x\alpha x_m$

Since $d_2>d_1$, the above proportionality equation of both the experiments can be given as:

$x_{m2}>x_{m1}$

Hence, the amplitude of second experiment is greater than the first experiment.

Calculation of the period of the experiments 

b)

From equation (iii), we can see that m and k are same for both systems, so period will not change, that is given by,${\mathrm{T}}_1={\mathrm{T}}_2$

Hence, the period remains the same in both experiments.

Calculation of the frequency of the experiments 

c)

From equation (iv), we can see that the period value is inversely proportional to frequency.

Again, as the period is same in both experiments, the frequency will also be the same.

Calculation of the maximum kinetic energy of the experiments 

d)

Kinetic energy is directly proportional to displacement from equation (v) and displacement is proportional to amplitude from part (a) calculations.

We have,$d_2>d_1$.Thus, we get the kinetic energies as:

$\mathrm{K}.{\mathrm{E}}_{\max 2}>\mathrm{K}.{\mathrm{E}}_{\max 1}$

Hence, the maximum kinetic energy of second experiment is greater than the first experiment.

Calculation of the maximum potential energy of the experiments 

e)

Potential energy is also directly proportional to displacement from equation (vi)

We have,$d_2>d_1$.Therefore, we get the potential energies as:

$U_{max2}>U_{max1}$

Hence, the maximum potential energy of second experiment is greater than the first experiment.