JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 2)
Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :
10 N
7.5 N
5 N
12.5 N
설명
y = 0.03 sin(450 t $$-$$ 9x)
v = $${\omega \over k} = {{450} \over 9}$$ = 50m/s
v = $$\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }$$ = 2500
$$ \Rightarrow $$ T = 2500 $$ \times $$ 5 $$ \times $$ 10$$-$$3
= 12.5 N
v = $${\omega \over k} = {{450} \over 9}$$ = 50m/s
v = $$\sqrt {{T \over \mu }} \Rightarrow {T \over \mu }$$ = 2500
$$ \Rightarrow $$ T = 2500 $$ \times $$ 5 $$ \times $$ 10$$-$$3
= 12.5 N