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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations5 Marks
Question
A spring having with a spring constant $1200 N m ^{-1}$ is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.
Image
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass. and (iii) the maximum speed of the mass.
✓

Answer

Given :
$
k=1200 N / m, m=3.0 kg
$
Maximum displacement
$
A=2.0 cm=2 \times 10^{-2} m
$
(i) We know time period
$
T=2 \pi \sqrt{\frac{m}{k}}
$
Frequency $\quad n=\frac{1}{T}$
$\begin{aligned} n & =\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\ \text { Put value } \quad & =\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}} \\ & =\frac{20}{2 \times 3.142}=3.18 \\ n & =3.18 \text { per second } \\ & =3.2 \text { per second }\end{aligned}$
(ii) Acceleration $a=-\omega^2 A=\frac{-k}{m} A$
$
\left|a_{\max }\right|=\frac{k}{m}\left|A_{\max }\right|
$
Here denoted $\omega=\sqrt{\frac{k}{m}}$
When displacement is maximum then acceleration will also be maximum
$
A=0.02 m
$
$a=\frac{1200}{3} \times 0.02=8.0 m / s ^2$
(iii) Keeping the speed value of maximum mass when it will pass through the
$\begin{aligned} v _{\max } & = A \omega= A \sqrt{\frac{k}{m}} \\ & =0.02 \times \sqrt{\frac{1200}{3}}=0.02 \sqrt{400} \\ & =0.02 \times 20=0.40 m / s .\end{aligned}$

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