## A particle is executing SHM according to the equation x = Acos ωt . Average speed of the particle during the interval …

Q: A particle is executing SHM according to the equation x = Acos ωt . Average speed of the particle during the interval 0 ≤ t ≤ π/6ω

(A) $\frac{\sqrt{3}\omega A}{2}$

(B) $\frac{\sqrt{3}\omega A}{4}$

(C) $\frac{3 \omega A}{\pi}$

(D) $\frac{3 \omega A}{\pi}(2-\sqrt{3})$

Click to See Solution :
Ans: (D)
Sol: Average Speed , $v_{avg} = \frac{\int_{0}^{t}\frac{dx}{dt}dt}{t}$

$v_{avg} = \frac{\int_{0}^{t}dx}{t}$

$= \frac{x(t)-x(0)}{t}$

$= \frac{A(cos\pi/6 – 1)}{\pi/6 \omega}$

$= \frac{3 \omega A}{\pi}(\sqrt{3} – 2 )$

since particle does not change it’s direction in the given interval , average speed

$v_{avg} = \frac{3 \omega A}{\pi}(2-\sqrt{3})$

## A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle ….

Q: A particle performs S.H.M. on x-axis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is:

(A) $\frac{T}{20}$

(B) $\frac{T}{2\pi} cos^{-1}(\frac{4}{5})$

(C) $\frac{T}{2\pi} cos^{-1}(\frac{1}{5})$

(D) $\frac{T}{2\pi} sin^{-1}(\frac{1}{5})$

Click to See Solution :
Ans: (B)
Sol: Particle is starting from rest, i.e. from one of its extreme position. As particle moves a distance A/5, we can represent it on a circle as shown

$\displaystyle cos\theta = \frac{4A/5}{A} = \frac{4}{5}$

$\displaystyle \theta = cos^{-1}(\frac{4}{5})$

$\displaystyle \omega t = cos^{-1}(\frac{4}{5})$

$\displaystyle t = \frac{1}{\omega} cos^{-1}(\frac{4}{5})$

$\displaystyle t = \frac{T}{2 \pi} cos^{-1}(\frac{4}{5})$

## A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration …

Q: A simple pendulum 50 cm long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration √3 g m/s2 .. The period of small oscillations of the pendulum about its equilibrium position is (g = π2m/s2) : (A) 1.0 sec

(B) 2 sec

(C) 1.53 sec

(D) 1.68 sec

Click to See Solution :
Ans: (A)
Sol: With respect to the cart, equilibrium position of the pendulum is shown.
If displaced by small angle θ from this position, then it will execute SHM about this equilibrium position, time period of which is given by : $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$

$T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$

$g_{eff} = \sqrt{g^2 + (\sqrt{3}g)^2}$

$g_{eff} = 2 g$

$T = 2\pi \sqrt{\frac{l}{g_{eff}}}$

## Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical…..

Q: Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The
pendulum bob has mass 0.2 kg. The length of the pendulum is equal to (g = 10 m/s^2 ) (A) 2.0 m

(B) 1.8 m

(C) 1.5 m

(D) 1.2

Click to See Solution :
Ans: (C)
Sol: $\frac{1}{2}m v_m^2 = 15 \times 10^{-3}$

$v_m = \sqrt{0.150}$

$\omega A = \sqrt{0.150} m/s$

$\sqrt{\frac{g}{L}} A = \sqrt{0.150} m/s$

L = 1.5 m