A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position ${\mathit{y}}_{\mathbf{i}}$such that the spring is at its rest length. The object is then released from ${\mathit{y}}_{\mathbf{i}}$and oscillates up and down, with its lowest position being $\mathbf{10}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{below}\mathbf{}{\mathit{y}}_{\mathbf{i}}$

(a) What is the frequency of the oscillation?

(b) What is the speed of the object when it is $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$below the initial position?

(c) An object of mass $\mathbf{300}\mathbf{}\mathit{g}$is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object?

(d) How far below ${\mathit{y}}_{\mathbf{i}}$is the new equilibrium (rest) position with both objects attached to the spring?

The given distance is twice the amplitude so$x=0.05\mathrm{m}$

Using the concepts of SHM and conditions given inthe problem, we can find all theanswers step by step. We can use the formula for frequency in terms of force constant and mass. To find velocity, we can use the law of conservation of energy. Using the new condition for mass, we can write the equations for frequency and compare it with the original equation for frequency to find the new frequency.

Formula:

Force acting on a body due to its weight,$F=mg\dots \dots \dots .\left(\mathrm{i}\right)$

Force on a spring,$F=Kx\dots \dots \dots \dots \left(ii\right)$

Frequency of oscillation,role="math" localid="1655099698569" $f=\frac{1}{2\left(\pi \right)}\sqrt{\frac{k}{m}}\dots \dots \dots \dots \left(\mathrm{iii}\right)$

Potential energy of a body,$PE=\frac{1}{2}k{x}^2\dots \dots \dots ..\left(\mathrm{iv}\right)$

Kinetic energy of a body,$KE=\frac{1}{2}m{v}^2\dots \dots \dots ..\left(\mathrm{v}\right)$

From equations (i) & (ii), we get

$$\begin{gathered} mg=kx \\ 9.8m=0.05k \\ \frac{k}{m}=196 \end{gathered}$$

Now, using equation (iii) and the above value, we can get the frequency as:

$$\begin{gathered} f=\frac{1}{2\left(3.14\right)}\sqrt{196} \\ =2.23\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency is$2.23Hz$

Using equation (iv), we can get the potential energy as:

$$\begin{gathered} P{E}_1=\frac{1}{2}k{\left(0.05\right)}^2 \\ =k\left(1.25\times {10}^{-3}\right) \end{gathered}$$

Now, potential energy at the position below 8 cm can be given as:

For this, the amplitude is $\left(0.08\mathrm{m}-0.05\mathrm{m}\right)=0.03\mathrm{m}$

So, the potential energy at that position

$$\begin{gathered} P{E}_2=\frac{1}{2}k{\left(0.03\right)}^2 \\ =k\left(4.5\times {10}^{-4}\right) \\ =k\left(4.5\times {10}^{-3}\right) \end{gathered}$$

Now, potential energy difference in both positions is given as:

$$\begin{gathered} ∆PE=\left(P{E}_1-P{E}_2\right) \\ =\left(0.8\times {10}^{-3}\right)\times k \end{gathered}$$

We know that energy is conserved during motion.

So we can write that the kinetic energy of a body is equal to that of the potential energy difference of that system. Hence, using equations (iv) & (v) and from potential energy equation, we get

$$\begin{gathered} \left(0.8\times {10}^{-3}\right)k=\frac{1}{2}m{v}^2 \\ {v}^2=\left(1.6\times {10}^{-3}\right)\frac{k}{m} \\ =\left(1.6\times {10}^{-3}\right)196 \\ v=0.56\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the object is$0.56\mathrm{m}/\mathrm{s}$

As for the condition in the given data that the system frequency is half that of the original frequency of the system with only mass m, we can say that

$$\begin{gathered} \frac{1}{2\pi }\sqrt{\frac{k}{m+∆m}}=\left(\frac{1}{2}\right)\left(\frac{1}{2\pi }\right)\sqrt{\frac{k}{m}} \\ \frac{k}{m+∆m}=\frac{1}{4}\left(\frac{k}{m}\right) \\ m+∆m=4m \\ m=0.100\mathrm{kg} \end{gathered}$$

Hence, the mass of the first object is $0.100\mathrm{kg}$$0.100\mathrm{kg}$

If both mass are attached to object, the mass of system becomes 0.4 kg, which is 4 times the original one.

Again, by comparing equations (i) & (ii), we get

$mg=kx$

Where g andk are constant

If we compare these masses, we can find the new equilibrium position

$$\begin{gathered} \frac{0.1g}{0.4g}=\frac{0.5k}{kx} \\ =0.200\mathrm{m} \end{gathered}$$

So, new equilibrium is at$0.2m$distance.