In the figure given below, a block of mass M = 490 g placed on a frictionless table is connected with two springs having same spring constant (K=2Nm -1 ). If the block is horizontally displaced through 'X'm then the number of complete oscillations it will make in 14n seconds will be________

Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(20)

Keff = K + K as both springs are in use in parallel
= 2k = 2x2 = 4N/m
m = 490gm = 0.49 kg


No. of oscillation in the
is

Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A block of mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The blo…
A mass m attached to free end of a spring executes SHJV t with a period of Is. If the mass is incre…
A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (-25 x) N. Th…
The general displacement of a simple harmonic oscillator is x-Asin t. Let T be its time period. Th…
The velocity of a particle executing SHM varies with displacement (x) as 4v 2 = 50 - x 2 . The time…
The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy wil…