A particle is subjected to two mutually perpendicular SHM such that x = 2sin$\mathrm{\omega }$t and

$\mathrm{y} = 2 \sin \left[\mathrm{\omega t} + \frac{\mathrm{\pi }}{2}\right]$. The path of the particle will be

1.  An ellipse

2.  A straight line

3.  A parabola

4.  A circle

When a periodic force \(\vec{F_1}\) acts on a particle, the particle oscillates according to the equation \(x=A\sin\omega t\). Under the effect of another periodic force \(\vec{F_2}\), the particle oscillates according to the equation \(y=B\sin(\omega t+\frac{\pi}{2})\). The amplitude of oscillation when the force (\(\vec{F_1}+\vec{F_2}\)) acts are:

A particle is executing SHM with amplitude A and angular frequency $\mathrm{\omega }$. The time taken by the particle to move from x = 0 to x = A/2 is

1.  $\frac{\mathrm{\pi }}{12 \mathrm{\omega }}$

2.  $\frac{\mathrm{\pi }}{6 \mathrm{\omega }}$

3.  $\frac{\mathrm{\pi }}{3 \mathrm{\omega }}$

4.  $\frac{\mathrm{\pi }}{\mathrm{\omega }}$

A spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)

A particle executes simple harmonic oscillations under the effect of small damping. If the amplitude of oscillation becomes half of the initial value of 16 mm in five minutes, then what will be the amplitude after fifteen minutes?

1.  8 mm

2.  4 mm

3.  2 mm

4.  1 mm

Which of the following is incorrect about simple harmonic oscillations of a particle? 

For a particle executing SHM, the graph between its momentum and displacement will be

1. Straight line

2. Hyperbola

3. Parabola

4. Ellipse

A particle executes SHM of amplitude 10 m and period 4 s along a straight line. The velocity of the particle at a distance of 6 m from the mean position is: 

1.  2$\mathrm{\pi }$ m/s

2.  4$\mathrm{\pi }$ m/s 

3.  6$\mathrm{\pi }$ m/s

4.  5.4$\mathrm{\pi }$ m/s

A uniform rod of mass m and length L pivoted from its one end is executing SHM with time period T. If rod suddenly breaks from the middle while passing through its mean position, then the time period of oscillation of remaining half part will be

1.  2T

2.  $\frac{\mathrm{T}}{2}$

3.  $\sqrt{2}\mathrm{T}$

4.  $\frac{\mathrm{T}}{\sqrt{2}}$