NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T . If the mass is increased by m , the time period become 5 T 3 . Then the ratio of m M is
Options
- A3 / 5
- B25 / 9
- C16 / 9
- D5 / 3
Correct answer
C. 16 / 9
Step-by-step solution
T = 2 π M k when mass is increased by m then ...(i) T = 2 π M + m k ⇒ 5 T 3 = 2 π M + m k ...(ii) Dividing Eq. (i) by Eq. (ii), we get 3 5 = M M + m 9 25 = M M + m ⇒ 9 M + 9 m = 25 M ⇒ 16 M = 9 m m M = 16 9