Question

Amplitude of vibrational of a particle in SHM with time period 1 sec is 0.05 m.average speed of the particle in one period is...
1) 0.4m/s. 2) 0.2m/s. 3) 0.1m/s. 4) 0.00m/s.

Answer 1

Answer on Question #51583-Physics-Mechanics-Kinematics-Dynamics

Amplitude of vibrational of a particle in SHM with time period $T = 1$ sec is $S = 0.05\,m$. average speed of the particle in one period is...

1) $0.4\,\frac{m}{s}$. 2) $0.2\,\frac{m}{s}$. 3) $0.1\,\frac{m}{s}$. 4) $0.00\,\frac{m}{s}$.

Solution

An average speed of the particle in one period is always zero!

But we can prove this.

Let the displacement of a particle in SHM be

$$x = A \cos(\omega t - \varphi),$$

where $A$, the maximum value of the displacement, is called the amplitude of the motion. If $T$ is the time for one complete oscillation and $\varphi$ is the phase angle, then the velocity $v$ is

$$v = \frac{dx}{dt} = -A\omega \sin(\omega t - \varphi) = -A\omega \sqrt{1 - \frac{x^2}{A^2}}.$$

The acceleration of the particle is

$$a = \frac{dv}{dt} = -A\omega^2 \cos(\omega t - \varphi) = -\omega^2 x.$$

Average speed of the particle in one period is

$$\tilde{v} = \frac{1}{T} \int_{t_0}^{t_0 + T} v(t) \, dt = \int_{t_0}^{t_0 + T} (-A\omega \sin(\omega t - \varphi)) \, dt = \frac{1}{T} \left(x(t_0 + T) - x(t_0)\right).$$

But $x(t_0 + T) = x(t_0)$ for periodic motion. Thus,

$$\tilde{v} = \frac{1}{T} \cdot 0 = 0.$$

Answer: 4) $0.00\,\frac{m}{s}$.

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