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Physics Question 36 – JEE-MAIN 2026

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A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are _______ Hz and _______ J, respectively. (Take g=10 m/s2)

First, determine the spring constant using the static equilibrium condition. Then, identify the amplitude of oscillation.

Step 1: Determine Spring Constant and Oscillation Parameters✦ Active

The spring stretches by xst=2 mm=2×10−3 m when loaded with mass m=200 g=0.2 kg. From the equilibrium position, the mass is further pulled down by A=2 mm=2×10−3 m, which is the amplitude of oscillation. The spring constant k is found from the static equilibrium condition mg=kxst.

k=mgxst=0.2 kg×10 m/s22×10−3 m=22×10−3=1000 N/m
Step 2: Calculate the Frequency of Oscillation○ Expand

The angular frequency ω and frequency f of a spring-mass system are given by:

ω=km=1000 N/m0.2 kg=5000=1050 rad/s

Therefore, the frequency f is:

f=ω2π=10502π=550π Hz
Step 3: Calculate the Maximum Energy in the Spring○ Expand

The maximum extension of the spring from its natural (unstretched) length during oscillation occurs at the lowest point. This maximum extension xmax is the sum of the static extension and the amplitude of oscillation.

xmax=xst+A=2×10−3 m+2×10−3 m=4×10−3 m

The maximum potential energy stored in the spring is given by:

Emax=12kxmax2=12×1000 N/m×(4×10−3 m)2

Calculating the value:

Emax=500×(16×10−6)=8000×10−6=8×10−3 J

Comparing the calculated frequency f=550π Hz and maximum energy Emax=8×10−3 J with the given options, Option 1 matches.

💡 Teacher's Secret Hint

Be careful to distinguish between the amplitude of oscillation and the maximum extension from the natural length when calculating energy.

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