A stretched string lies along the x-axis. The string is displaced along both the y-and z- direction, so that the transverse displacement of the string is given by
y(x, t) = A cos(kx – ωt) z(x, t) = A sin(kx – ωt)
Text Solution
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y 2 (x, y) + z 2 (x, y) = A 2
The trajectory is a circle of radius A.
At t = 0, y(0, 0) = A, z(0, 0) = 0.
At t = π /2 ω , y(0, π /2 ω ) = 0, z(0, π /2 ω ) = – A.
At t = π / ω , y(0, π / ω ) = – A, z(0, π /2 ω ) = 0.
At t = 3 π /2 ω ,y(0, 3 π /2 ω ) = 0, z(0, 3 π /2 ω ) = +A.
v y = dy/dt = +A ω sin(kx – ω t), v z = dz/dt = –A ω cos(kx – ω t)
v = 
= A ω , so the speed is constant.
= y 
+ z 
. 
= yv y + zv z = A 2 ω sin(kx – ω t)cos(kx – ω t) – A 2 ω cos(kx – ω t) sin(kx – ω t)
. 
= 0, so 
is tangent to the circular path.
a y = dv y /dt = –A ω 2 cos(kx – ω t) a z = dv z /dt = –A ω 2 sin(kx – ω t)
. 
= ya y + za z = A 2 ω 2 [cos 2 (kx – ω t) + sin 2 (kx – ω t)] = – A 2 ω 2
r = A, a = A ω 2 , so 
. 
= –ra
. 
= ra cos φ so φ = 180º and 
is opposite in direction to 
; 
is radially inward.
y 2 + z 2 = A 2 , so the path is again circular, but the particle rotates in the opposite sense compared to part .
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