SHMHard
Question
A mass m is suspended from a spring of length l and force constant K. The frequency of vibration of mass is f1. The spring is cut into two equal parts and the some mass is suspended from one of the parts. The new frequency of vibration of mass is f2. Which of the following relations between the frequencies is correct :-
Options
A.f1 = √2 f2
B.f1 = f2
C.f1 = 2f2
D.f2 = √2f1
Solution
When spring is cut into two equal parts then spring constant of each part will be 2K and
f ∝ √K
New frequency will be √2 times so f2 = √2f1
f ∝ √K
New frequency will be √2 times so f2 = √2f1
Create a free account to view solution
View Solution FreeMore SHM Questions
A thin rod of length 1 m is suspended from its end and is made to oscillate in a vertical plane. The distance between th...Values of the acceleration of a particle moving in simple harmonic motion as a function of its displacement x are given ...The maximum tension in the string of a pendulum is three times of the minimum tension. If θ0 be the angular amplitu...If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is :...The equation of motion of a particle of mass 1 g is + π2x = 0 where x is displacement (in m) from mean position. Th...