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NVAEasyJEE 2023Simple Harmonic Motion (SHM)

JEE Physics 2023 Question with Solution

In the figure given below, a block of mass M=490 gM = 490 \, g is placed on a frictionless table is connected with two springs having the same spring constant (K=2 N/mK = 2 \, N/m). If the block is horizontally displaced through XX m, then the number of complete oscillations it will make in 14t14t seconds will be:

A block labeled M placed on a horizontal frictionless table between two identical springs labeled K, each attached to rigid walls on both sides.

Answer

Correct answer:20

Step-by-step solution

Standard Method

Given: Mass of the block is m=490 g=0.49 kgm = 490 \, g = 0.49 \, \text{kg} and each spring has spring constant K=2 N/mK = 2 \, \text{N/m}.

Find: Number of complete oscillations in the given time interval.

A block between two springs fixed to opposite walls, each spring labeled K = 2 N/m, showing the horizontal oscillation setup.

Since both springs act together, the effective spring constant is

Keff=K+K=2K=2×2=4 N/mK_{\text{eff}} = K + K = 2K = 2 \times 2 = 4 \, \text{N/m}

The time period of oscillation is

T=2πmKeffT = 2\pi \sqrt{\frac{m}{K_{\text{eff}}}}

Substituting the values,

T=2π0.494=2π49400=2π×720=7π10 sT = 2\pi \sqrt{\frac{0.49}{4}} = 2\pi \sqrt{\frac{49}{400}} = 2\pi \times \frac{7}{20} = \frac{7\pi}{10} \, \text{s}

The number of oscillations in the given time interval is

N=timeT=14π7π/10=20N = \frac{\text{time}}{T} = \frac{14\pi}{7\pi/10} = 20

Therefore, the number of complete oscillations is 2020.

Using the extracted working

Given: Two identical springs of constant K=2 N/mK = 2 \, \text{N/m} are attached to a block of mass 490 g490 \, g.

Find: The number of complete oscillations made in the stated time.

When two springs are connected in parallel, their spring constants add:

Keff=K+K=2k=4 N/mK_{\text{eff}} = K + K = 2k = 4 \, \text{N/m}

Also,

m=490 g=0.49 kgm = 490 \, g = 0.49 \, \text{kg}

Now,

T=2πmKeff=2π0.494=2π40049−1=2π⋅720=7π10T = 2\pi \sqrt{\frac{m}{K_{\text{eff}}}} = 2\pi \sqrt{\frac{0.49}{4}} = 2\pi \sqrt{\frac{400}{49}}^{-1} = 2\pi \cdot \frac{7}{20} = \frac{7\pi}{10}

Using the solution conclusion,

N=14π7π/10=20N = \frac{14\pi}{7\pi/10} = 20

So the final numerical answer is 2020.

Common mistakes

  • Treating the two springs as if they were in series. That is wrong because the block moves with both springs acting simultaneously, so the effective spring constant is the sum. Use Keff=K+KK_{\text{eff}} = K + K.

  • Forgetting to convert mass from grams to kilograms. Using 490490 instead of 0.49 kg0.49 \, \text{kg} gives an incorrect time period. Always convert to SI units before substitution.

  • Using amplitude XX in the time period formula. For ideal SHM of a spring-block system, the time period does not depend on amplitude. Use only mm and KeffK_{\text{eff}}.

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