A standing wave on a string has a frequency of 100 Hz, a wavelength of 0.04 m, and an amplitude of 2 mm. The transverse velocity of the point x = 0.02 m, is:
Solution.
ν=100Hz;\nu=100Hz;ν=100Hz;
λ=0.04m;\lambda=0.04m;λ=0.04m;
A=0.002m;A=0.002m;A=0.002m;
v−?;v-?;v−?;
v=ωA;v=\omega A;v=ωA;
ω=2πν=2⋅3.14⋅100=628s−1;\omega=2\pi\nu=2\sdot3.14\sdot100=628s^{-1};ω=2πν=2⋅3.14⋅100=628s−1;
v=628⋅0.002=1.256m/s;v=628\sdot0.002=1.256 m/s;v=628⋅0.002=1.256m/s;
Answer:v=1.256m/s.v=1.256 m/s.v=1.256m/s.
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