Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
c. More intense IR absorption occur for those bonds having greater dipole moment changes with bond lengthening in a vibration.
Explanation:
When the molecules is exposed to the infrared radiation, the sample molecules absorb the radiation of wavelengths (specific to molecule) which causes change in the dipole moment of the sample molecules. The vibrational energy levels of the sample molecules consequently transfer from the ground state to the excited state.
Frequency of absorption peak is determined by vibrational energy gap.
Intensity of absorption peaks is related to change of dipole moment and possibility of transition of the energy levels.
Thus, by analyzing infrared spectrum,abundant structure information of the molecule can be known.
Hence, the correct answer to the question is
c. More intense IR absorption occur for those bonds having greater dipole moment changes with bond lengthening in a vibration.
The answer is square route of pie and william