91Ó°ÊÓ

The given air-track glider, which is mounted on a spring oscillates between the $10cm$ mark and the $60cm$ mark on the track.

$T=$Time period.

$t=$Time.

$n=$No. of oscillations.

Then,$T=\frac{t}{n}$.

Since the glider completes $10$oscillations in $33s$, $t=33s$and $n=10$.

Substitute them into the formula to obtain $T$.

$$\begin{gathered} T=\frac{t}{n} \\ =\frac{33s}{10} \\ =3.3s \end{gathered}$$

The time period is$3.3s$.

The given air-track glider, which is mounted on a spring oscillates between the $10cm$ mark and the $60cm$ mark on the track.

Frequency$=\frac{1}{timeperiod}$.

The time period is $3.3s$.

Hence, the frequency is

role="math" localid="1648523592892" $$\begin{gathered} f=\frac{1}{timeperiod} \\ =\frac{1}{3.3s} \\ =0.303Hz. \end{gathered}$$

The air-track glider has a frequency of$0.303Hz$.

The air-track glider has a frequency of $\frac{1}{3.3s}=0.303Hz$.

If $f$is the frequency then, the angular frequency of air-track glider is$\mathrm{Ó\neg }=2\mathrm{Ï€´Ú}$.

$$\begin{gathered} \mathrm{Ó\neg }=2\mathrm{Ï€´Ú} \\ \mathrm{Ó\neg }=2×3.14×0.303Hz \\ \mathrm{Ó\neg }=1.902rad/s \end{gathered}$$

The air-track glider has a angular frequency of$1.902rad/s$.

The given air-track glider, which is mounted on a spring oscillates between the $10cm$ mark and the $60cm$ mark on the track.

The given air-track glider travel a distance $=60cm-10cm=50cm$.

The amplitude of the given air-track glider is$\frac{50cm}{2}=25cm$.

The amplitude of the given air-track glider is$25cm$.

The air-track glider has a angular frequency $\mathrm{Ó\neg }=1.902rad/s$.

The amplitude of the given air-track glider$a=25cm$.

Maximum speed, $v_{max}=\mathrm{Ó\neg }a$

Where,$\mathrm{Ó\neg }=$angular frequency, and$a=$amplitude.

The maximum speed of the given air-track glider is

$$\begin{gathered} v_{max}=\mathrm{Ó\neg }a \\ =1.902ra/s×25cm \\ =47.55cm/s \\ =0.48m/s. \end{gathered}$$

The maximum speed of the given air-track glider is$0.48m/s$.