Amplitude (A), Period (T) and Frequency (n) of S.H.M.

5.6.1 Amplitude of S.H.M.

Diagram (Fig. 5.2): a particle P performing S.H.M. along a straight line MN, with the centre O of MN as its mean position. M and N are the two extreme positions, each at a distance $A$ from O.

The displacement is $x=A\sin(\omega t+\phi)$. The particle has its maximum displacement when $\sin(\omega t+\phi)=\pm1$, i.e. when $x=\pm A$. This distance $A$ is called the amplitude.

Definition: The maximum displacement of a particle performing S.H.M. from its mean position is called the amplitude of S.H.M.

The total path length of the S.H.M. (from one extreme to the other) is $MN=2A$.

5.6.2 Period of S.H.M.

Definition: The time taken by the particle performing S.H.M. to complete one oscillation is called the period of S.H.M.

The displacement at time $t$ is $x=A\sin(\omega t+\phi)$. After a time $t'=t+\dfrac{2\pi}{\omega}$ the displacement is $$x=A\sin\left[\omega\left(t+\dfrac{2\pi}{\omega}\right)+\phi\right]=A\sin(\omega t+2\pi+\phi)=A\sin(\omega t+\phi)$$

The particle is at the same position after a time $\dfrac{2\pi}{\omega}$, and it can be shown that this is the minimum such time. Hence $$T=\dfrac{2\pi}{\omega}$$

Since $\omega^2=\dfrac{k}{m}=\dfrac{\text{force per unit displacement}}{\text{mass}}=$ acceleration per unit displacement, $$T=\dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}\qquad\text{and}\qquad T=2\pi\sqrt{\dfrac{m}{k}}$$

5.6.3 Frequency of S.H.M.

Definition: The number of oscillations performed by a particle performing S.H.M. per unit time is called the frequency of S.H.M.

In time $T$ the particle performs one oscillation, so in unit time it performs $\dfrac{1}{T}$ oscillations: $$n=\dfrac{1}{T}=\dfrac{\omega}{2\pi}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$

Combination of springs

Springs of different spring constants can be combined in series, in parallel, or both.

Series combination

Diagram (Fig. A): the springs are connected one after the other, forming a single chain, with the stretching force $f$ applied at the free end.

• If the springs are massless, each carries the same stretching force $f$ (for a vertical arrangement, the weight $mg$).
• If $e_1$ and $e_2$ are the extensions, $f=k_1e_1=k_2e_2$, so $e_1=\dfrac{f}{k_1}$ and $e_2=\dfrac{f}{k_2}$.
• Total extension: $e=e_1+e_2=f\left(\dfrac{1}{k_1}+\dfrac{1}{k_2}\right)$.
• If $k_s$ is the effective spring constant, $e=\dfrac{f}{k_s}$, hence

$$\dfrac{1}{k_s}=\dfrac{1}{k_1}+\dfrac{1}{k_2}\qquad\Rightarrow\qquad k_s=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{\text{Product}}{\text{Sum}}$$

• For a number of such springs in series, $\dfrac{1}{k_s}=\sum\dfrac{1}{k_i}$; for $n$ identical springs of constant $k$, $k_s=\dfrac{k}{n}$.

Parallel combination

Diagram (Fig. B): all the springs are connected between the same two points — one of them the support, the other the point at which the stretching force $f$ is applied.

• Irrespective of their spring constants, each spring now has the same extension $e$.
• The springs share the force so that in the equilibrium position the total restoring force is equal and opposite to $f$: $f=f_1+f_2+\ldots=k_1e+k_2e+\ldots$
• If $k_p$ is the effective spring constant, $f=k_pe$, hence

$$k_p=k_1+k_2+\ldots=\sum k_i$$

• For $m$ identical massless springs of constant $k$ in parallel, $k_p=mk$.

For two identical springs of constant $k$: $k_s=\dfrac{k}{2}$ and $k_p=2k$, so since $n\propto\sqrt{k}$ the frequencies are in the ratio $n_s:n_p=1:2$.

Standard results

• $A=x_{max}$; path length $=2A$
• $T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{m}{k}}=\dfrac{2\pi}{\sqrt{\text{acceleration per unit displacement}}}$
• $n=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$
• Series: $\dfrac{1}{k_s}=\sum\dfrac{1}{k_i}$, $k_s=\dfrac{k}{n}$ for $n$ identical springs
• Parallel: $k_p=\sum k_i$, $k_p=mk$ for $m$ identical springs