The displacement of a particle undergoing SHM of time period T is given by xt - xmcos - t - - The particle is at x - -xm at time t - 0- The particle is at x - -xm when

The correct option is B t=0.50T The time taken by particle to move from left extreme #160;to right extreme is equal to half of the time #160;period. i ...

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Byju's Answer

Standard XII

Physics

Phase and the General Equation

The displacem...

Question

The displacement of a particle undergoing $S H M$ of time period $T$ is given by $x (t) = x_{m} c o s (ω t - ϕ)$ . The particle is at $x = - x_{m}$ at time $t = 0$ . The particle is at $x = + x_{m}$ when

t = 0.25 T

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t = 0.50 T

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t = 0.75 T

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t = 1.00 T

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Solution

The correct option is B $t = 0.50 T$
The time taken by particle to move from left extreme to right extreme is equal to half of the time period. i.e. $0.5 T$

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The displacement of a particle undergoing SHM of time period T is given by x(t)=xmcos(ωt−ϕ). The particle is at x=−xm at time t=0. The particle is at x=+xm when

The correct option is B t=0.50TThe time taken by particle to move from left extreme to right extreme is equal to half of the time period. i.e. 0.5T

The displacement of a particle undergoing $S H M$ of time period $T$ is given by $x (t) = x_{m} c o s (ω t - ϕ)$ . The particle is at $x = - x_{m}$ at time $t = 0$ . The particle is at $x = + x_{m}$ when

The correct option is B $t = 0.50 T$
The time taken by particle to move from left extreme to right extreme is equal to half of the time period. i.e. $0.5 T$