question_answer

Figure shows a sinusoidal wave of period T travelling to the right along a string at time t = 0. Which of the following statement is incorrect?

A) The point 3 on the string is moving upward with maximum speed

B) The point 5 on the string has the greatest upward acceleration

C) The point 9 on the string has the greatest downward acceleration

D) The point 2 on the string has a downward velocity and upward acceleration

Correct Answer: D

Solution :

Displacement \[y\,\,(x,t)=A\cos \,\,(kx-\omega t)\] At \[t=0,\] point 1 on the string has maximum displacement.

Velocity \[{{v}_{y}}(x,t)=\omega A\,\,\,sin(kx-\omega t)\]

At \[t=0,\]\[{{v}_{y}}\]is maximum at \[kx=\frac{\pi }{2}\]

\[\Rightarrow x=\frac{\pi \lambda }{2\times 2\pi }=\frac{\lambda }{4}\] i.e., point 3 on the string.

Hence [A] is correct.

Acceleration \[{{a}_{y}}(x,t){{\omega }^{2}}A\,\,\,\cos (kx-\omega t)\]At \[t=0,\] \[{{a}_{y}}\]is maximum downward acceleration

If \[\cos kx=1\Rightarrow kx=2\pi \Rightarrow x=\lambda \]

i.e., point 9 on the string has maximum downward acceleration. Hence [C] is correct.

At t=0, \[kx=-\pi ,\] \[x=\frac{\lambda }{2}\] i.e., the string has maximum upward acceleration.

Hence [B] is correct

Velocity \[{{v}_{y}}\] is positive and acceleration \[{{a}_{y}}\] is negative when \[0<kx<\frac{\pi }{2}\]

\[{{v}_{y}}\] is negative and \[{{a}_{y}}\] is positive \[\pi <kx<\frac{3\pi }{2}\]

\[{{v}_{y}}\] and \[{{a}_{y}}\] are positive when \[\frac{\pi }{2}<kx<\pi \]

\[{{v}_{y}}\] and \[{{a}_{y}}\] are negative when \[\frac{3\pi }{2}<kx<\pi \] [D] is incorrect.