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Worked Examples · Example 13.7

Q.A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m−150\ \text{N m}^{-1}. The block is pulled to a distance x=10x = 10 cm from its equilibrium position at x=0x = 0 on a frictionless surface from rest at t=0t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.

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In SHM, total mechanical energy is constant and given by 12kA2\frac{1}{2}kA^2. At x=5x = 5 cm, potential energy is 12kx2\frac{1}{2}kx^2 and kinetic energy is the difference between total and potential energy. The values are: U=0.0625 JU = 0.0625\ \text{J}, K=0.1875 JK = 0.1875\ \text{J}, Etotal=0.25 JE_{\text{total}} = 0.25\ \text{J}.

The key insight here is that in Simple Harmonic Motion on a frictionless surface, mechanical energy is conserved. The spring force is conservative, so the sum of kinetic and potential energy never changes. Once you know the amplitude, you know the total energy — and from there, finding the split at any position is just a matter of plugging in.

Let’s walk through it.

  1. Identify the amplitude and total energy. The block is pulled to x=10 cm=0.1 mx = 10\ \text{cm} = 0.1\ \text{m} and released from rest. That’s the maximum displacement — the amplitude AA. At x=Ax = A, the block is momentarily at rest, so all energy is potential:

Etotal=12kA2E_{\text{total}} = \frac{1}{2} k A^2

Plug in k=50 N/mk = 50\ \text{N/m} and A=0.1 mA = 0.1\ \text{m}:

Etotal=12×50×(0.1)2=12×50×0.01=0.25 JE_{\text{total}} = \frac{1}{2} \times 50 \times (0.1)^2 = \frac{1}{2} \times 50 \times 0.01 = 0.25\ \text{J}

Etotal=12kA2E_{\text{total}} = \frac{1}{2} k A^2

  1. Find the potential energy at x=5 cmx = 5\ \text{cm}. At any displacement xx, the spring potential energy is:

U=12kx2U = \frac{1}{2} k x^2

Here x=5 cm=0.05 mx = 5\ \text{cm} = 0.05\ \text{m}:

U=12×50×(0.05)2=12×50×0.0025=0.0625 JU = \frac{1}{2} \times 50 \times (0.05)^2 = \frac{1}{2} \times 50 \times 0.0025 = 0.0625\ \text{J}

  1. Find the kinetic energy at that position. Since total energy is constant:

K=Etotal−U=0.25−0.0625=0.1875 JK = E_{\text{total}} - U = 0.25 - 0.0625 = 0.1875\ \text{J}

Watch out

A common mistake is to forget that xx must be in metres, not centimetres. Using x=5x = 5 instead of 0.050.05 gives U=6250 JU = 6250\ \text{J} — wildly wrong. Always convert cm to m before plugging into formulas with N/m\text{N/m}. …

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