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Simple Harmonic Motion question

2023 · 29 Jan · Shift 2 · Q65
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  5. /2023 · 29 Jan · Shift 2 · Q65

Simple Harmonic Motion question

2023 · 29 Jan · Shift 2 · Q65

JEE MainPhysicsSimple Harmonic MotionNumerical+4 / −1
A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(−25 x)N\mathrm{F}=(-25~x)\mathrm{N}F=(−25 x)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
View written solutionFree

Correct answer: 40

  1. Identify the restoring force form

    In SHM, restoring force is F=−kxF=-kxF=−kx Comparing with the given force F=−25xF=-25xF=−25x we get k=25 N/mk=25\ \text{N/m}k=25 N/m

  2. Write the mass

    Given mass: m=250 g=0.25 kgm=250\ \text{g}=0.25\ \text{kg}m=250 g=0.25 kg

  3. Find angular frequency

    For SHM, ω=km\omega=\sqrt{\frac{k}{m}}ω=mk​​ Substituting values, ω=250.25=100=10 rad/s\omega=\sqrt{\frac{25}{0.25}}=\sqrt{100}=10\ \text{rad/s}ω=0.2525​​=100​=10 rad/s

  4. Use relation between maximum speed and amplitude

    In SHM, maximum speed is vmax⁡=ωAv_{\max}=\omega Avmax​=ωA Given vmax⁡=4 m/sv_{\max}=4\ \text{m/s}vmax​=4 m/s Hence, A=vmax⁡ω=410=0.4 mA=\frac{v_{\max}}{\omega} = \frac{4}{10}=0.4\ \text{m}A=ωvmax​​=104​=0.4 m

  5. Convert into cm

    0.4 m=40 cm0.4\ \text{m}=40\ \text{cm}0.4 m=40 cm

Final Answer

The amplitude of motion is 40 cm\boxed{40\ \text{cm}}40 cm​

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