A simple pendulum has a time period \(T_1\) when on the earth’s surface, and \(T_2\) when taken to a height \(R\) above the earth’s surface, where \(R\) is the radius of the earth. The value of \(\frac{T_2}{T_1}\) is:
1. \(1\)
2. \(\sqrt{2}\)
3. \(4\)
4. \(2\)

A particle executes linear simple harmonic motion with an amplitude of of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity  is equal to that of its acceleration. Then, its time period in seconds is 

(a) $\sqrt{\frac{5}{\pi }}$

(b)$\sqrt{\frac{5}{2\mathrm{\pi }}}$

(c)$\frac{4\mathrm{\pi }}{\sqrt{5}}$

(d)$\frac{2\mathrm{\pi }}{\sqrt{3}}$

A body mass m is attached to the lower end of a spring whose upper end is fixed. The spring has neglible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. The value of m in kg is-

(1) $\frac{3}{4}$               

(2)$\frac{4}{3}$

(3) $\frac{16}{9}$               

(4) $\frac{9}{16}$

When two displacements represented by y₁=asin(ωt) and y₂=bcos(ωt) are superimposed,the motion is -

(1) not a simple harmonic

(2) simple harmonic with amplitude a/b

(3) simple harmonic with amplitude $\sqrt{a^{2 }+ b^2}$

(4) simple harmonic with amplitude (a+b)/2

A partricle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is  β. Then, its time period of vibration will be

(1)β²/α² 

(2)α/β

(3)β²/α

(4)2πβ/α 

The damping force on an oscillator is directly proportional to the velocity.The units of the constant of proportionality are 

(1)$kg m{s}^{-1}$                               

(2)$kg m{s}^{-2}$

(3)$kg s^{-1}$                                   

(4)$kg s$

The period of oscillation of a mass \(M\) suspended from a spring of negligible mass is \(T.\) If along with it another mass \(M\) is also suspended, the period of oscillation will now be:
1. \(T\)
2. \(T/\sqrt{2}\)
3. \(2T\)
4. \(\sqrt{2} T\)