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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2026 · 2 April · Evening Shift · Physics · Oscillations

The equation of motion of a particle is given by x = a (50t +...

The equation of motion of a particle is given by x=asin⁡x = a \sin\left(50t + \dfrac{\pi}{3}\right)cm.Theparticlewillcometorestattimecm. The particle will come to rest at timet_1 and it will have zero acceleration at time t2t_2. The t1t_1 and t2t_2 respectively are _______.

Options

  1. Aπ300 s,π75 s\dfrac{\pi}{300} \, s, \dfrac{\pi}{75} \, s
  2. Bπ75 s,π300 s\dfrac{\pi}{75} \, s, \dfrac{\pi}{300} \, s
  3. Cπ300 s,π25 s\dfrac{\pi}{300} \, s, \dfrac{\pi}{25} \, s
  4. Dπ50 s,π100 s\dfrac{\pi}{50} \, s, \dfrac{\pi}{100} \, s

Correct answer

A. π300 s,π75 s\dfrac{\pi}{300} \, s, \dfrac{\pi}{75} \, s

Step-by-step solution

The equation of motion is given by x=asin⁡(50t+π3)x = a \sin\left(50t + \dfrac{\pi}{3}\right).

Velocity is the rate of change of displacement:
v=dxdt=50acos⁡(50t+π3)v = \dfrac{dx}{dt} = 50a \cos\left(50t + \dfrac{\pi}{3}\right)

The particle comes to rest when v=0v = 0. For the first time it comes to rest:
cos⁡(50t1+π3)=0\cos\left(50t_1 + \dfrac{\pi}{3}\right) = 0

50t1+π3=π250t_1 + \dfrac{\pi}{3} = \dfrac{\pi}{2}

50t1=π2−π3=π650t_1 = \dfrac{\pi}{2} - \dfrac{\pi}{3} = \dfrac{\pi}{6}

t1=π300 st_1 = \dfrac{\pi}{300} \text{ s}

Acceleration is the rate of change of velocity:
A=dvdt=−2500asin⁡(50t+π3)A = \dfrac{dv}{dt} = -2500a \sin\left(50t + \dfrac{\pi}{3}\right)

The particle has zero acceleration when A=0A = 0. For the first time it has zero acceleration:
sin⁡(50t2+π3)=0\sin\left(50t_2 + \dfrac{\pi}{3}\right) = 0

50t2+π3=π50t_2 + \dfrac{\pi}{3} = \pi

50t2=π−π3=2π350t_2 = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}

t2=2π150=π75 st_2 = \dfrac{2\pi}{150} = \dfrac{\pi}{75} \text{ s}

Answer: π300 s,π75 s\dfrac{\pi}{300} \, s, \dfrac{\pi}{75} \, s
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