A mass m is attached to two springs as shown in figure. The spring constants of two springs are K 1 and K 2 . For the frictionless surface, the time period of oscillation of mass m is

Question

Aicharya · Accepted Answer

For the given combination, if the block is displaced towards right, elongation in one spring will be equal to the compression in other and the direction of force due to both spring will be the same and restoring in nature. Hence, both springs are effectively in parallel combination.
Therefore,
$\Rightarrow K_{e f f} = K_{1} + K_{2}$ and
$\Rightarrow T = 2 π \sqrt{\frac{m}{K_{e f f}}} = 2 π \sqrt{\frac{m}{K_{1} + K_{2}}}$

Alternative Solution: Bilkul samjhate hai, dhyaan se suno. Dekho, jab hum ek mass ko two springs ke saath attach karte hai, toh humein iska time period of oscillation nikalna hota hai. Iska formula hota hai $T = 2\pi \sqrt{\frac{m}{k}}$, jahan $m$ mass hai aur $k$ spring constant hai. Ab yahaan pe humare paas do springs hai, jinki spring constants hai $K_1$ aur $K_2$. Jab do ya do se zyada springs ek saath hote hain, toh unka equivalent spring constant hota hai un sabka sum. Matlab, $K_{eq} = K_1 + K_2$. Ab hum yeh $K_{eq}$ wali value apne formula mein daalenge. Toh humara formula banega $T = 2\pi \sqrt{\frac{m}{K_{eq}}}$ yaani $T = 2\pi \sqrt{\frac{m}{K_1 + K_2}}$. Toh bilkul sahi ja rahe ho, bas aise hi concepts ko samjho, practice karte raho, aur JEE ki taiyari mast chalegi. Samjha?

A mass m is attached to two springs as shown in figure. The spring constants of two springs are K 1 and K 2 …

Solution