An oscillating block–spring system has a mechanical energy of 1.00 J, amplitude of 1.00 cm, and a maximum speed of 1.20 m/ s.

1. Find the spring constant.
2. Find the mass of the block.
3. Find the frequency of oscillation.

1. The amplitude of motion,${\mathrm{X}}_{\mathrm{m}}=10.0\mathrm{cm}\mathrm{or}0.10\mathrm{m}.$.
2. The maximum speed of block,$v_m=1.20\mathrm{m}/\mathrm{s}$.
3. The mechanical energy of system, E = 1.0 J.

We can find the spring constant and the mass of the block using the law of conservation of energy. Then we can find the frequency of oscillation of the block by using the formula for the frequency of SHM.

Formulae:

The elastic potential energy of the system$P.E=\frac{1}{2}{\mathrm{kx}}^2,\dots \dots \dots \left(1\right)$

The kinetic energy of the system $K.E=\frac{1}{2}{\mathrm{mv}}^2......\left(2\right)$

The law of conservation of energy gives .........(3)

The frequency of oscillation for SHM$f=\frac{1}{2\mathrm{\tau \tau }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}..........\left(4\right)$

From equation (iii), we can say that the mechanical energy of system is equal to the maximum elastic P.E energy of system. Hence, the total energy of the system is given as:

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\mathrm{kx}}_{\mathrm{m}}^2(\because \mathrm{from}\mathrm{equation}(1\left)\right) \\ \mathrm{K}=\frac{2\mathrm{E}}{{\mathrm{x}}_{\mathrm{m}}^2} \\ =\frac{2\left(1\right)}{{(0.10)}^2} \\ =200\mathrm{N}/\mathrm{m} \end{gathered}$$

Therefore, the spring constant is 200 N / s.

Similarly, from equation (iii), we can say that the mechanical energy of system = the maximum elastic K.E energy of system. Hence, the total energy of the system is given as:

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{m}}^2(\because \mathrm{from}\mathrm{equation}(2\left)\right) \\ \mathrm{m}=\frac{2\mathrm{E}}{{{\mathrm{v}}_{\mathrm{m}}}^2} \\ =\frac{2\left(1\right)}{{1.2}^2} \\ =1.39\mathrm{Kg} \end{gathered}$$

Therefore, the mass of the block is 1.39 Kg.

The frequency of oscillation of block using equation (iii)and the given values is given as:

$$\begin{gathered} \mathrm{f}=\frac{1}{2\mathrm{\tau \tau }}\sqrt{\frac{200}{1.39}} \\ =1.91\mathrm{Hz} \end{gathered}$$

Therefore, the frequency of oscillation of the block is 1.91Hz