On a frictionless, horizontal air track, a glider oscillates at the end of an ideal spring of force constant 2.50 N/cm. The graph in Fig. E14.19 shows the acceleration of the glider as a function of time. Find (a) the mass of the glider; (b) the maximum displacement of the glider from the equilibrium point; (c) the maximum force the spring exerts on the glider.

Any spring that is frictionless, massless and linear is said to be Ideal spring. Ideal spring obeys Hooke’s law.

Mathematically it is given by,

F = - kx

Here, F is the force due to spring

k is the spring constant of spring

x is the displacement

Force constant of spring,$k=2.50\mathrm{N}/\mathrm{cm}=2.50\frac{\mathrm{N}}{\mathrm{cm}}\times \frac{100\mathrm{cm}}{1\mathrm{m}}=250\mathrm{N}/\mathrm{m}$

From the given figure,

Time,$T=0.30s-0.10s=0.20s$

Maximum acceleration,$a_{max}=\pm 12\mathrm{m}/{\mathrm{s}}^2$

(a) The frequency of SHM of a glider is given by,

$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{m}}$ …(1)

The relation between frequency and periodic time is given by,

$f=\frac{1}{T}$ …(2)

From equation. (1) and (2),

$\frac{1}{T}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{m}}$

Rewrite the above expressions for

$m=\frac{k{T}^2}{4{\mathrm{\pi }}^2}$

Substitute the given values in above expression to get the mass of glider

$$\begin{gathered} m=\frac{\left(250N/m\right){\left(0.20s\right)}^2}{4{\mathrm{\pi }}^2} \\ =0.253kg \end{gathered}$$

Therefore, mass of the glider is 0.253 kg.

(b)

It is known that the maximum displacement of the glider is the amplitude of SHM and the spring force is at its maximum value when the glider reaches its maximum displacement. This implies that the acceleration of the glider is also at its maximum.

Apply Newton’s second law in the x-direction,

$F=m{a}_x$ …(3)

The force due to spring having spring constant is given by,

$F=-kx$ …(4)

From equation. (3) and (4),

$-kx=m{a}_x$ …(5)

At the maximum displacement of the glider,$x=A$ and $a_x=a_{max}$. Therefore, equation. (5) can be written as

$\begin{array}{l}-kA=m{a}_{max}\\ \therefore A=-\frac{m{a}_{max}}{k}\end{array}$

Substitute values in above expression to get the maximum displacement of the glider from the equilibrium position

$\begin{array}{r}A=-\frac{\left(0.253\mathrm{kg}\right)\left(-12\mathrm{m}/{\mathrm{s}}^2\right)}{250\mathrm{N}/\mathrm{m}}\\ =0.0121\mathrm{m}\\ =1.21\mathrm{cm}\end{array}$

Therefore, maximum displacement of the glider from the equilibrium position is 1.21 cm.

The maximum force exerted on the glider by the spring is given by,

$F_{max}=m{a}_{max}$

Substitute values in above expression

$\begin{array}{r}{F}_{max}=\left(0.253\mathrm{kg}\right)\left(12\mathrm{m}/{\mathrm{s}}^2\right)\\ =3.04\mathrm{N}\end{array}$

Therefore, maximum force the spring exerts on the glider is 3.04 N.