A body of mass m is attached to the lower end of a spring whose upper end is fixed- The spring has negligible mass- When the mass m is slightly pulled down and released- it oscillates with a time period of 3s- when the mass m is increased by 1 kg the time period of oscillations becomes 5s- Then value of m in kg isdfrac-9-16-dfrac-3-4-dfrac-16-9-dfrac-4-3-

Question

Toppr Admin · Accepted Answer

For SHM of a hanging mass by a spring-F-x2212-ksxwhere ks-m-x3C9-2-x27F9-x3C9-x221A-ksmHence time period of SHM-T-2-x3C0-x3C9-2-x3C0-x221A-mks-x221D-x221A-mHence 53-x221A-m-1m-x27F9-m-916kg

For SHM of a hanging mass by a spring,F=−ksxwhere ks=mω2⟹ω=√ksmHence time period of SHM=T=2πω=2π√mks∝√mHence 53=√m+1m⟹m=916kg

For SHM of a hanging mass by a spring,
$F = - k_{s} x$
where $k_{s} = m ω^{2}$
$⟹ ω = \sqrt{\frac{k_{s}}{m}}$
Hence time period of SHM= $T = \frac{2 π}{ω}$
$= 2 π \sqrt{\frac{m}{k_{s}}} \propto \sqrt{m}$
Hence $\frac{5}{3} = \sqrt{\frac{m + 1}{m}}$
$⟹ m = \frac{9}{16} k g$