A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of wate

Q: Why do the vibrations stop?

The water provides a viscous drag force (damping) that opposes the motion of the mass, gradually draining its mechanical energy.

Q: Why do we ignore the kinetic energy of the mass?

The spring is stretched and *released* from rest. At the moment of release, its kinetic energy is zero, and all mechanical energy is stored as potential energy in the spring.

Q: Why do we add the heat capacities of both the mass and the water?

Because they are in thermal contact. When the mechanical energy turns into heat, it warms up the entire system (mass + water) together to a new equilibrium temperature.

Q: What if the container wasn't perfectly insulated?

If the container absorbed heat, we would need to add its heat capacity ($m_cs_c$) to our equation. Since it's negligible here, we ignore it.

Q: Is the order of magnitude always the nearest power of 10?

Yes, $3.65 \times 10^{-5}$ is closer to $10^{-5}$ than to $10^{-4}$ in logarithmic scale, so the order of magnitude is $10^{-5}$.

Question

A massless spring ($k = 800 	ext{ N/m}$), attached with a mass ($500 	ext{ g}$) is completely immersed in $1 	ext{ kg}$ of water. The spring is stretched by $2 	ext{ cm}$ and released, so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely?
(Assume that the water container and spring receive negligible heat and specific heat of mass $= 400 	ext{ J/kg K}$, specific heat of water $= 4184 	ext{ J/kg K}$)

Options:

(A) $10^{-4} 	ext{ K}$

(B) $10^{-3} 	ext{ K}$

(C) $10^{-1} 	ext{ K}$

(D) $10^{-5} 	ext{ K}$

Sigma Percentile · Accepted Answer

### The Setup: A Submerged Oscillator

Imagine a block of mass $m$ attached to a spring of constant $k$, completely submerged in a container of water. We stretch the spring by a distance $x_m$ and let it go.

It begins to oscillate, but this isn't an ideal, frictionless world. The water provides a viscous drag, a damping force that opposes the motion of the block.

### The Fate of Mechanical Energy

Because of this damping, the amplitude of the oscillations gradually decreases until the block comes to a complete stop. But where does the energy go?

The First Law of Thermodynamics tells us that energy cannot be destroyed. The initial mechanical energy of the system, which was entirely stored as elastic potential energy in the stretched spring, is dissipated as heat.

This heat doesn't just vanish; it is absorbed by the water and the block, causing their temperature to rise.

### Calculating the Energy

Let's quantify this. The initial mechanical energy $E_{	ext{mech}}$ is given by the potential energy of the stretched spring:

$E_{	ext{mech}} = \frac{1}{2} k x_m^2$

Substituting the given values ($k = 800 	ext{ N/m}$ and $x_m = 2 	ext{ cm} = 0.02 	ext{ m}$):

$E_{	ext{mech}} = \frac{1}{2} (800) (0.02)^2 = 0.16 	ext{ J}$

### The Thermal Response

Now, this $0.16 	ext{ J}$ of heat energy is absorbed by the block and the water. To find the temperature rise $\Delta T$, we need the total heat capacity $C$ of the system.

The heat capacity is the sum of the heat capacities of the individual components:

$C = m_{	ext{block}} s_{	ext{block}} + m_{	ext{water}} s_{	ext{water}}$

Plugging in the masses and specific heats:

$C = (0.5 	ext{ kg})(400 	ext{ J/kg K}) + (1 	ext{ kg})(4184 	ext{ J/kg K})$

$C = 200 + 4184 = 4384 	ext{ J/K}$

### The Final Temperature Rise

Equating the mechanical energy to the thermal energy ($Q = C \Delta T$):

$0.16 = 4384 	imes \Delta T$

$\Delta T = \frac{0.16}{4384} \approx 3.65 	imes 10^{-5} 	ext{ K}$

This is an incredibly tiny temperature rise! Looking at our options, the order of magnitude is clearly **$10^{-5} 	ext{ K}$**.

This problem beautifully illustrates how macroscopic mechanical energy transforms into microscopic thermal energy, bridging the gap between mechanics and thermodynamics.

Select Answer:

Visualized Solution

System Setup

Conservation of Energy

Initial Mechanical Energy

Total Heat Capacity

Equating Energies

Final Calculation

The Way Forward

The Sigma Insight: Free, Forced, and Damped Oscillations

The Setup

The Fate of Mechanical Energy

Calculating the Energy

The Thermal Response

The Final Temperature Rise

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