1.3 Linear shm  (Page 2/4)

• If displacement is sine function, then velocity function is cosine function and vice-versa.
• The range of velocity lies between “-ωA” and “ωA”.
• The velocity attains maximum value two times in a cycle at the center – (i) moving from negative to positive extreme and then (ii) moving from positive to negative extreme.
• The velocity at extreme positions is zero.

Acceleration

The acceleration in linear motion is given as :

$$a=-{\omega }^{2}x=-\frac{k}{m}x$$

Substituting for displacement “x”, we get an expression in variable time, “t” :

$$\Rightarrow a=-{\omega }^{2}x=-{\omega }^{2}A\sin \left(\omega t+\phi \right)$$

We can obtain this relation also by differentiating displacement function twice or by differentiating velocity function once with respect to time. Few important points about the nature of acceleration should be kept in mind :

1: Acceleration changes its direction about point of oscillation. It is always directed towards the center whatever be the position of the particle executing SHM.

2: Acceleration linearly varies with negative of displacement. We have seen that force-displacement plot is a straight line. Hence, acceleration – displacement plot is also a straight line. It is positive when “x” is negative and it is negative when “x” is positive.

3: Nature of force with respect to time, however, is not linear. If we combine the expression of acceleration and displacement, then we have :

$$\Rightarrow a=-{\omega }^{2}x=-{\omega }^{2}A\sin \left(\omega t+\phi \right)$$

Here, we draw both displacement and acceleration plots with respect to time in order to compare how acceleration varies as particle is at different positions.

The upper figure is displacement – time plot, whereas lower figure is acceleration – time plot. We observe following important points about variation of acceleration :

• If displacement is sine function, then acceleration function is also sine function, but with a negative sign.
• The range of acceleration lies between “ $-{\omega }^{2}A$ ” and “ ${\omega }^{2}A$ ”.
• The acceleration attains maximum value at the extremes.
• The acceleration at the center is zero.

4: Since force is equal to product of mass and acceleration, F = ma, it is imperative that nature of force is similar to that of acceleration. It is given by :

$$\Rightarrow F=ma=-kx=-m{\omega }^{2}x=-m{\omega }^{2}A\sin \left(\omega t+\phi \right)$$

Frequency, angular frequency and time period

The angular frequency is given by :

$$\omega =\sqrt{\left(\frac{k}{m}\right)}=\sqrt{\left|\frac{a}{x}\right|}=\sqrt{\left|\frac{\text{acceleration}}{\text{displacement}}\right|}$$

We have used the fact " $F=ma=-kx$ " to write different relations as above.

Time period is obtained from the defining relation :

$$T=\frac{2\pi }{\omega }=2\pi \sqrt{\left(\frac{m}{k}\right)}=2\pi \sqrt{\left|\frac{x}{a}\right|}=\sqrt{\left|\frac{\text{displacement}}{\text{acceleration}}\right|}$$

Frequency is obtained from the defining relation :

$$\nu =\frac{1}{T}=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\left(\frac{k}{m}\right)}=\frac{1}{2\pi }\sqrt{\left|\frac{a}{x}\right|}=\sqrt{\left|\frac{\text{acceleration}}{\text{displacement}}\right|}$$

Kinetic energy

The instantaneous kinetic energy of oscillating particle is obtained from the defining equation of kinetic energy as :

$$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k\left(A^2-x^2\right)$$

The maximum value of KE corresponds to position when speed has maximum value. At x = 0,

$$\Rightarrow K_{\text{max}}=\frac{1}{2}m{\omega }^2\left(A^2-0^2\right)=\frac{1}{2}k{A}^2$$

The minimum value of KE corresponds to position when speed has minimum value. At x = A,