A particle starts SHM from the mean position. Its amplitude is a and total energy E. At one instant its kinetic energy is 3 · Its displacement at that instant is
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
B.
When the particle of mass m at O is pushed by y in the direction of A, the spring A will be compressed by y while spring B and C will be stretched by y' = y cos 45°. So, that the total restoring force on the mass m is along OA
Fnet = FA + FB cos 45° + FC cos 45°
= ky + 2ky' cos 45°
= ky + 2k (y cos 45°) cos 45°
= 2 ky
Also, Fnet = k'y ⇒ k'y = 2ky ⇒ k' = 2k
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