Question

A massless spring attached to a rigid support at one end and a mass m at other end is placed on smooth surface. The mass is pulled and released. It executes SHM with time period T. If time period for given systems is given by T'. Then match the following. (Given that in part d energy gets lost, but consider only spring force for calculating time period)

• A. A-s, b-q, c-p, d-r
• B. A-s, b-p, c-q, d-r
• C. A-r, b-q, c-p, d-r
• D. A-s, b-q, c-r, d-p

Answer 1

The correct option is A A-s, b-q, c-p, d-r

$\left(a\right)\to \left(s\right)\left(b\right)\to \left(q\right);\left(c\right)\to \left(p\right);\left(d\right)\to \left(r\right)$


$\therefore T=2\pi \sqrt{\frac{m}{k_{eq}}}$


$T^{\prime }=2\pi \sqrt{\frac{m}{k_{eq}}}=2\pi \sqrt{\frac{2m}{k}}=\sqrt{2}T$
$(\text{As}\frac{1}{k_{eq}}=\frac{2}{k}\therefore k_{eq}=\frac{k}{2})$


$T^{\prime }=2\pi \sqrt{\frac{m}{k_{eq}}}=2\pi \sqrt{\frac{m}{2k}}=\frac{T}{\sqrt{2}}$


$T^{\prime }=2\pi \sqrt{\frac{m}{k}}=T\left(\text{undamped}\right)$


$T^{\prime }=2\pi \sqrt{\frac{m}{k}}=T\text{(damped)}$

Why this question? Tip: Remember the formulae for finding $k_{eff}$ in series and parallel combinations of spring to save the effort and time of using force method to find time period.