1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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 becomes 5T/3. then the ratio of m/M is

A
3/5
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
25/9
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
16/9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5/3
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C 16/9Time period of oscillation of the block executing SHM T=2π√MKwhere M and K is the mass of the block and spring constant of the spring.Thus new time period T′=5T3=2π√M+mK ∴5T3T=√M+mMSquaring both sides 259=1+mM⟹mM=169

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Simple Pendulum
PHYSICS
Watch in App
Join BYJU'S Learning Program