The time period of simple harmonic motion of mass M in the given figure is M5k, where the value of is . [figure]
12
Right branch: two k in parallel =2k, in series with k: 2k k2k+k= 2k3
Left spring k is in parallel with this branch: keq=k+ 2k3= 5k3
T=2 Mkeq=2 3M5k= 12M5k
=12
Answer: 12
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be s. [take g= 2 m s-2]
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8, where x= .
A simple harmonic oscillator has an amplitude A and time period 6 second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=A to x= 32 A will be x s, where x= .
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is cm, where = .