A ball of mass $^{'} m^{'}$ kg hangs from a spring of force constant $K$ and current length $^{'} l^{'}$ . The ball oscillates with a period of $T$ seconds. If the ball removed, the spring is shortened ( w.r.t. length in mean position) by:

\frac{g T^{2}}{(2 π)^{2}}

meter

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 T^{2} g}{(2 π)^{2}}

meter

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{T m}{k}

meter

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{T k}{m}

meter

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{g T^{2}}{(2 π)^{2}}$ meter
Time period when mass m hangs from spring
$T = 2 π \sqrt{\frac{m}{k}} m g = k x x = \frac{m g}{k} = \frac{m g \times T^{2}}{{(2 π)}^{2} m} = \frac{g T^{2}}{{(2 π)}^{2}}$

Suggest Corrections

Similar questions

A ball of mass 2 kg hanging from a spring oscillates with a time period $2 π$ seconds. Ball is removed when it is in equilibrium position, then spring shortens by

Q. A ball of mass

2 k g

hanging from a spring oscillates with a time period

2 s

. Ball is removed when it is in equilibrium position, then spring shortens by:

Q. A body of mass 5 kg hangs from a spring and oscillates with a time period of

2 π

second. If the body is removed, the length of the spring will decrease by

Q. An iron ball of mass

M

is hanged from the ceiling by a spring with a spring constant

k

. It executes a

S H M

with a period

P

. If the mass of the ball is increased by four times, the new period will be

Q. A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in

\frac{T}{3}

seconds

Join BYJU'S Learning Program

A ball of mass ′m′ kg hangs from a spring of force constant K and current length ′l′. The ball oscillates with a period of T seconds. If the ball removed, the spring is shortened ( w.r.t. length in mean position) by:

The correct option is A gT2(2π)2 meterTime period when mass m hangs from springT=2π√mkmg=kxx=mgk=mg×T2(2π)2m=gT2(2π)2

A ball of mass $^{'} m^{'}$ kg hangs from a spring of force constant $K$ and current length $^{'} l^{'}$ . The ball oscillates with a period of $T$ seconds. If the ball removed, the spring is shortened ( w.r.t. length in mean position) by:

The correct option is A $\frac{g T^{2}}{(2 π)^{2}}$ meter
Time period when mass m hangs from spring
$T = 2 π \sqrt{\frac{m}{k}} m g = k x x = \frac{m g}{k} = \frac{m g \times T^{2}}{{(2 π)}^{2} m} = \frac{g T^{2}}{{(2 π)}^{2}}$