Oscillations Te...

Maximum tension to the minimum tension in the string during oscillations is?(Consider amplitude of oscillation =$$ \theta $$)

A particle rests in equilibrium under two forces of repulsion whose centres are at a distance of $$a$$ and $$b$$ from the particle. The forces vary as the cube of the distance. The forces per unit mass are $$k$$ and $$k'$$ respectively. If the particle is slightly displaced towards one of them then, the motion is simple harmonic with the time period equal to:

The period of a conical pendulum of string length $$\sqrt 2$$ is $$2 \ s$$. The radius of the bob's orbit is about horizontal is $$[g=9.8 \ m/s^2]$$

A particle executing SHM while moving from one extreme is found at distance $$x_1,\, x_2$$ and $$x_3$$ from the centre at the end of three successive seconds. The time period of oscillation is: Here  $$\theta  = {\cos ^{ - 1}}\left( {\frac{{{x_1} + {x_3}}}{{2{x_2}}}} \right)$$

Three measurements of the time for $$20$$ oscillations of a pendulum give $$t_1 = 39.6_s , t_2 = 39.9_s and t_3 = 39.5_s$$. The precision in the measurement, is  

A particle of mass 0.1 kg executes SHM under a force F= (-10x)N. Speed of particle at mean position is 6 m/s. Then amplitude of oscillations is

If two wires of same length $$\ell $$ and area of cross section A with young modulus y and 2y connect in series and one end is fixed on roof and other end with mass m. Make simple harmonic motion, then the time period is-

A linear harmonic oscillator of force constant $$6 \times 10^5 N/m$$ and amplitude $$4 $$ cm, has total energy $$600 J$$. Select the correct statement.

Equation of SHM is x = 10 sin 10 $$\pi t$$. Find the distance between the two points where speed is $$50\pi$$ m/s. x is in cm and t is in seconds. 

A  particle executes SHM  along the line AB . IFC divided AB in the ratio 3 : 1, the ration of the time taken to travel AC and CB is :