Laws of MotionHard
Question
A 20 cm long string having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.
Options
A.2.5 cm
B.5 cm
C.7.5 cm
D.10 cm
Solution
L = 20 cm
m = 1 gm
μ =
gm / cm = 
g / m
μ =
kg / m
T = 0.5 N v =
= 10 m/s
f = 100 Hz
λ =
m
m = 5 cm
m = 1 gm
μ =
gm / cm = g / mμ =
kg / mT = 0.5 N v =
= 10 m/sf = 100 Hz
λ =
mm = 5 cmCreate a free account to view solution
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