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प्रश्न
The following figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
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उत्तर
Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = –1.2 s)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = – ω2x ...(i)
where ω is the angular frequency and x is the displacement from the equilibrium position.
The velocity is the derivative of displacement with respect to time `v = dx/dt` ...(ii)
At time t = 0.3 s, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative.However, according to equation (i), the acceleration of the particle is positive.
At time t = 1.2 s, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
At time t = –1.2 s, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, according to equation (ii), the velocity is positive. As the position is negative, according to equation (i), the acceleration is positive.
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