SHM in Vertical Spring Block
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In the given figure, a mass is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is . The mass oscillates on a frictionless surface with time period and amplitude . When the mass is in an equilibrium position, as shown in the figure, another mass is gently fixed upon it. The new amplitude of oscillation will be:
Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
k1k2
√k1k2
k2k1
√k2k1
- 2mgK
- mgK
- 3mgK
- mg2K
[Assume the top surface of the block (represented by line AB) always remains horizontal]
- ω=√4k3M
- ω=√8k3M
- ω=√2k3M
- ω=√k3M
- Remain unchanged
- become 6f
- become f6
- become √6f
- mgsinθK
- 3mgsinθK
- gsinθ√Km
- 2mgsinθK
- 0.1πsec
- 0.5πsec
- π
- 2πsec
- T=2π(m(2k1+4k2+2k3))12
- T=2π(m(1k1+4k2+1k3))12
- T=2π(m(2k1+2k2+2k3))12
- T=2π(m(1k1+1k2+1k3))12
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released suddenly with the spring initially unstretched. Then the maximum extension in the spring is
4Mgk
2Mgk
Mgk
Mg2k
[Assume the top surface of the block (represented by line AB) always remains horizontal]
- ω=√4k3M
- ω=√8k3M
- ω=√2k3M
- ω=√k3M
- T=2π√m2K
- T=2π√mK
- T=2π√2mK
- T=2π√3m2K
A block is resting on a piston which is moving vertically with SHM of period 1 s. What is the minimum amplitude to be given to it so that the block and piston can separate?
0.2 m
0.25 m
0.3 m
0.35 m
- 0.5 m/s
- 1 m/s
- 2 m/s
- 4m/s
- √3:√2
- 3: 2
- 3√3:2√2
- 9 : 4
Youngs modulus of a perfectly rigid body is
zero
unity
finite and low
infinite
A spring is loaded with two blocks m1 and m2 where m1 is rigidly fixed with the spring and m2 is just kept on the block m1 as shown in the figure. The maximum energy of oscillation that is possible for the system having the block m2 in contact with m1 is
m21g22k
m22g22k
(m1+m2)2g22k
(m1+m2)2g2k
Find the ratio of lengths RC and RD.
- 3:1
- 3:2
- 2:1
- 4:3
- 1.0 kg
- 1.6 kg
- 2.0 kg
- 2.4 kg
A spring block system with mass M and spring constant k is suspended vertically and left to oscillate. It has a natural frequency fo f1. Another spring block system with same mass and spring constant is kept on a horizontal smooth surface.Its natural frequency is observed to be f2.
What do you think will be the relation between f1 and f2.
- f1>f2
- f1<f2
- f1=f2
- Data Insufficient
A block of mass m moves with a speed v towards the right block in equilibrium with a spring. If the surface is frictionless and collistions are elastic, the frequency of collisions between the masses will be :
v2L+1π√Km
2 [v2L+1π√Km]
2[2Lv+1π√Km]
4[2Lv+1π√Km]
One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance x0 from the block. The spring is then compressed by 2x0 and released. The time taken to strike the wall is
16π√km
√km
2π3√mk
π4√km
- mg−4π2T2mA
- mg+4π2T2mA
- mg−π2T2mA
- mg+π2T2mA
- √3:√2
- 3: 2
- 3√3:2√2
- 9 : 4
- 20 N
- 40 N
- 60 N
- 80 N
A spring is loaded with two blocks m1 and m2 where m1 is rigidly fixed with the spring and m2 is just kept on the block m1 as shown in the figure. The maximum energy of oscillation that is possible for the system having the block m2 in contact with m1 is
m21g22k
m22g22k
(m1+m2)2g22k
(m1+m2)2g2k
- 12mA2ω2
- 12mA2ω2+12ky2
- 12ky2
- 12mA2ω2−12ky2
- √ka+kbm
- √kakb(ka+kb)m
- √kakb4m(ka+kb)
- √4kakb(ka+kb)m
- Mgk[1+π23]
- Mgk[1+π29]
- Mg2k[1+π23]
- Mg2k[1+π29]
- Block A will start SHM.
- Amplitude of oscillation of the block A is mgk.
- Maximum speed acquired by the block A is √mg2k.
- Period of oscillation is 2π√m2k.
- Mgk[1+π23]
- Mgk[1+π29]
- Mg2k[1+π23]
- Mg2k[1+π29]