JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 13)
Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is :
_25th_February_Evening_Shift_en_13_1.png)
$$2\pi \sqrt {{m \over k}} $$
$$\pi \sqrt {{m \over k}} $$
$$2\pi \sqrt {{m \over {2k}}} $$
$$\pi \sqrt {{m \over {2k}}} $$
Explanation
Due to parallel combination Keff = 2k + 2k = 4k
$$\because$$ $$T = 2\pi \sqrt {{m \over {{k_{eff}}}}} $$
$$ = 2\pi \sqrt {{m \over {4k}}} $$
$$T = \pi \sqrt {{m \over k}} $$
$$\because$$ $$T = 2\pi \sqrt {{m \over {{k_{eff}}}}} $$
$$ = 2\pi \sqrt {{m \over {4k}}} $$
$$T = \pi \sqrt {{m \over k}} $$
Comments (0)
