Other Examples of SHM
Trending Questions
[(Take m=5 kg and k=10 N/m]
- T=π√lμg
- T=π√l4μg
- T=2π√2lμg
- T=2π√lμg
- f=12π√3k27M+7m
- f=1π√3k27m+7m
- f=12π√k27M+7m
- f=1π√k27M+7m
A trolley of mass M oscillates with some time period about its mean position when it is connected with an ideal spring of stiffness constant K.
If an unknown mass m is gently placed at the top of trolley, its time period is found to be T.Calculate the coefficient of static friction so that the upper mass does not slip even for sizeable amplitude A.
KAmg
KA(M+m)g
KAmmg
None of these
A long uniform rod of length L and mass M is free to rotate in a vertical plane about a horizontal axis through its one end 'O'. A spring of force constant k is connected vertically between one end of the rod and ground. When the rod is in equilibrium it is parallel to the ground.
What is the period of small oscillations that result when the rod is rotated slightly and released?
T=2π√M3k
T=2π√Mk
T=2π√2Mk
T=2π√2M3k
- 2π√√2a6g
- 2π√2√2a3g
- 2π√2ag
- 2π√a2g
- T=2π√2m3k
- T=2π√m3k
- T=2π√7m3k
- T=2π√4m3k
A long uniform rod of length L and mass M is free to rotate in a vertical plane about a horizontal axis through its one end 'O'. A spring of force constant k is connected vertically between one end of the rod and ground. When the rod is in equilibrium it is parallel to the ground.
What is the period of small oscillations that result when the rod is rotated slightly and released?
T=2π√M3k
T=2π√Mk
T=2π√2Mk
T=2π√2M3k
- π√2ag
- 2π√2ag
- 3π√2ag
- 2π√ag
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R2 from the earth's center where R is the radius of the earth. the wall of the tunnel is frictionless. Find the time period.
2π√RM
- 2π√R3GM
√2R3GM
2π√MR
Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance R2 from the earth's center where R is the radius of the earth. the wall of the tunnel is frictionless. Find the time period.
2π√RM
- 2π√R3GM
√2R3GM
2π√MR
[Take l=1 m, b=12√2 m, k1=16 N/m, k2=1 N/m, m=164 kg]
- ω=√k21a2m(k1a2+k2l2)
- ω=√k22a2m(k1a2+k2l2)
- ω=√k1k2a2m(k1a2+k2l2)
- None of these
- 2π√2l3g
- 2π√4√2l3g
- 2π√2l3g
- 3π√l3g
[Force constant of the spring is k and mass of the fixed pulley is negligible]
- T=2π√M1+M2k
- T=2π√M1+4M2k
- T=2π√M2+4M1k
- T=2π√M2+3M1k
- 2π√Lgcosα
- 2π√Lgsinα
- 2π√Lg
- 2π√Lgtanα
- π
- π2
- π3
- π4
[Take l=1 m, b=12√2 m, k1=16 N/m, k2=1 N/m, m=164 kg]
- π
- π2
- π3
- π4