JEE Main 2021 · 26 February · morning · Single correct · +4 / −1
If two similar springs each of spring constant K₁ are joined in series, the new spring constant and time period would be changed by a factor :
- (a) 1/2,2√2
- (b) 1/4,2√2
- (c) 1/2,√2
- (d) 1/4,√2
Answer and solution
Answer (c) 1/2,√2
Solution
Physics sir · 4 steps
In series the springs share the stretch, so the combined spring is softer: half as stiff.
- Series combination. In series, the same force acts on both springs and their stretches add. So the reciprocals of the spring constants add: 1/K = 1/K₁ + 1/K₁ = 2/K₁.
- New spring constant. This gives K = K₁/2. The spring constant changes by a factor of 1/2.
T = 2π√(m/K) ⇒ T_(new)/T = √(K₁/(K₁/2)) = √2
- Time period. The same mass hangs on the new spring. Since T ∝ 1/√K, halving K multiplies T by √2.
- So. The spring constant changes by 1/2 and the time period by √2, which is option (c).
The idea to keep. Springs in series add like resistors in parallel: 1/K = 1/K₁ + 1/K₂. Then use T ∝ 1/√K, so the period changes by the square root of the inverse factor.
Answer from the NTA final answer key. Solution by PrepWiser.