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NEET PHYSICSEasy

If x=5sin⁡(πt+π3) mx = 5 \sin(\pi t + \frac{\pi}{3}) \text{ m} represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively are:

A

5 m,2 s5 \text{ m}, 2 \text{ s}

B

5 cm,1 s5 \text{ cm}, 1 \text{ s}

C

5 m,1 s5 \text{ m}, 1 \text{ s}

D

5 cm,2 s5 \text{ cm}, 2 \text{ s}

Step-by-Step Solution

  1. Standard Equation of SHM: The general displacement equation for a particle executing simple harmonic motion is given by x(t)=Asin⁡(ωt+ϕ)x(t) = A \sin(\omega t + \phi), where:
  • AA is the Amplitude (maximum displacement from the mean position).
  • ω\omega is the Angular Frequency.
  • ϕ\phi is the Phase Constant (or initial phase).
  1. Comparing Equations: We compare the given equation x=5sin⁡(πt+π3)x = 5 \sin(\pi t + \frac{\pi}{3}) with the standard form. The coefficient of the sine function corresponds to the amplitude: A=5A = 5. Since the displacement xx is given in meters (m), the amplitude is A=5 mA = 5 \text{ m}. The coefficient of time tt inside the sine argument corresponds to the angular frequency: ω=π rad/s\omega = \pi \text{ rad/s}.
  2. Calculate Time Period (TT): The time period TT is related to the angular frequency by the formula T=2πωT = \frac{2\pi}{\omega}.
  • Substituting the value of ω\omega: T=2ππ=2 sT = \frac{2\pi}{\pi} = 2 \text{ s}.
  1. Conclusion: The amplitude is 5 m5 \text{ m} and the time period is 2 s2 \text{ s}.
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