The angular velocities of three bodies in simple harmonic motion are ω1,ω2,ω3ω1,ω2,ω3 with their respective amplitudes as A1,A2,A3A1,A2,A3. If all the three bodies have same mass and maximum velocity, then
1. A1ω1=A2ω2=A3ω3A1ω1=A2ω2=A3ω3
2. A1ω12=A2ω22=A3A32A1ω12=A2ω22=A3A32
3. A12ω1=A22ω2=A32ω3A12ω1=A22ω2=A32ω3
4. A12ω12=A22ω22=A2A12ω12=A22ω22=A2
The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 8√3cm/sec8√3cm/sec will be
1. 2√3cm2√3cm
2. √3cm√3cm
3. 1 cm
4. 2 cm
The maximum velocity of a simple harmonic motion represented by y=3 sin (100t+π6)y=3 sin (100t+π6) is given by
1. 300
2. 3π63π6
3. 100
4. π6π6
The displacement equation of a particle is x=3sin 2t+4cos 2t x=3sin 2t+4cos 2t The amplitude and maximum velocity will be respectively
1. 5, 10
2. 3, 2
3. 4, 2
4. 3, 4
The instantaneous displacement of a simple pendulum oscillator is given by x=A cos (ωt+π4)x=A cos (ωt+π4) . Its speed will be maximum at time
1. π4ωπ4ω
2. π2ωπ2ω
3. πωπω
4. 2πω2πω
The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is
1 144π2m/sec2144π2m/sec2
2 144m/sec2144m/sec2
3. 144π2m/sec2144π2m/sec2
4. 288π2m/sec2288π2m/sec2
A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by
1. – A Kx
2. A cos (Kx)
3. A exp (– Kx)
4. A Kx
What is the maximum acceleration of the particle doing the SHM y=2sin[πt2+ϕ]y=2sin[πt2+ϕ] where 2 is in cm
1. π2cm/s2π2cm/s2
2. π22cm/s2π22cm/s2
3. π4cm/s2π4cm/s2
4. π4cm/s2π4cm/s2