A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration . Then its time period in second is
Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are k1 and k2 respectively. The period of vertical oscillations of the system will be
Two massless springs of force constants k1 and k2 are joined end to end. The resultant force constant k of the system is
A spring of force constant k is cut into two equal halves. The force constant of each half is
k
2k
A particle of mass m is attached to three identical massless springs of spring constant k as shown in the figure. The time period of vertical oscillation of the particle is
A spring of force constant k is cut into three equal parts. The force constant of each part would be
3k
k
2k
A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy of the particle half potential and half kinetic ?
Two identical springs are connected to mass m as shown (k = spring constant). If the period of the configuration in (a) is 2s, the period of the configuration in (b) is
1 s
A particle of mass m is located in a one dimensional potential field where potential energy is given by V(x) = A(l − cos px), where A and p are constants. The period of small oscillations of the particle is
B.
We are given that a particle of mass m is located in a one dimensional potential field and the potential energy is given by V(x) = A(1 − cos px). So, we can find the force experienced by the particle as
For small oscillations, we have
F ≈ − Ap2x
Hence, the acceleration would be given by
Also we know that
The period of oscillation of a simple pendulum of length l suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α is given by