Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 12517 by tawa last updated on 24/Apr/17

A spring stretches by 15cm when a mass of 300g hangs down from it. if the spring is then strethed an additional 10cm and realeased, calculate (a) the spring constant (b) Angular velocity (c) The amplitude of the oscillation (d) The maximum velocity (e) The maximum acceleration of the mass (f) The period T and frequency f

$$\mathrm{A}\:\mathrm{spring}\:\mathrm{stretches}\:\mathrm{by}\:\mathrm{15cm}\:\mathrm{when}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{300g}\:\mathrm{hangs}\:\mathrm{down}\:\mathrm{from}\:\mathrm{it}. \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{is}\:\mathrm{then}\:\mathrm{strethed}\:\mathrm{an}\:\mathrm{additional}\:\mathrm{10cm}\:\mathrm{and}\:\mathrm{realeased},\:\mathrm{calculate} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{the}\:\mathrm{spring}\:\mathrm{constant} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Angular}\:\mathrm{velocity} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{The}\:\mathrm{amplitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{oscillation} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{The}\:\mathrm{maximum}\:\mathrm{velocity} \\ $$$$\left(\mathrm{e}\right)\:\mathrm{The}\:\mathrm{maximum}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mass} \\ $$$$\left(\mathrm{f}\right)\:\mathrm{The}\:\mathrm{period}\:\mathrm{T}\:\mathrm{and}\:\mathrm{frequency}\:\mathrm{f} \\ $$

Answered by mrW1 last updated on 24/Apr/17

(a) k=F/δ=mg/δ=0.3×10/0.15=20 N/m (b) w=(√(k/m))=(√((20)/(0.3)))=8.165 1/s (c) x_(max) =10 cm (d) v_(max) =w×x_(max) =8.165×0.1=0.816 m/s (e) a_(max) =w×v_(max) =8.165×0.816=6.662 m/s^2 (f) T=((2π)/w)=((2π)/(8.165))=0.729 sec f=(1/T)=1.371 Hz

$$\left({a}\right)\:{k}={F}/\delta={mg}/\delta=\mathrm{0}.\mathrm{3}×\mathrm{10}/\mathrm{0}.\mathrm{15}=\mathrm{20}\:{N}/{m} \\ $$$$\left({b}\right)\:{w}=\sqrt{\frac{{k}}{{m}}}=\sqrt{\frac{\mathrm{20}}{\mathrm{0}.\mathrm{3}}}=\mathrm{8}.\mathrm{165}\:\mathrm{1}/{s} \\ $$$$\left({c}\right)\:{x}_{{max}} =\mathrm{10}\:{cm} \\ $$$$\left({d}\right)\:{v}_{{max}} ={w}×{x}_{{max}} =\mathrm{8}.\mathrm{165}×\mathrm{0}.\mathrm{1}=\mathrm{0}.\mathrm{816}\:{m}/{s} \\ $$$$\left({e}\right)\:{a}_{{max}} ={w}×{v}_{{max}} =\mathrm{8}.\mathrm{165}×\mathrm{0}.\mathrm{816}=\mathrm{6}.\mathrm{662}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left({f}\right)\:{T}=\frac{\mathrm{2}\pi}{{w}}=\frac{\mathrm{2}\pi}{\mathrm{8}.\mathrm{165}}=\mathrm{0}.\mathrm{729}\:{sec} \\ $$$${f}=\frac{\mathrm{1}}{{T}}=\mathrm{1}.\mathrm{371}\:{Hz} \\ $$

Commented by tawa last updated on 24/Apr/17

God bless you sir. Thanks for everytime. i really appreciate

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{everytime}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com