A string of length 1.5 m with its two ends
clamped is vibrating in fundamental mode.
Amplitude at the centre of the string is
4 mm. Distance between the two points
having amplitude 2 mm is
(a) 1 m
(b) 75 cm
(c) 60 cm
(d) 50 cm
PLZZ answer for 50 points.
Answers
Answer:
The distance between the two points of a string whose length is 1.5 m is 1 meter.
Explanation:
Step 1 : write down the relevant equation.
Y = The amplitude of the wave.
Y = ASinKx × Cosbt
Now set t = 0 then we have :
Y = ASinkx
k = 2pie/lambda
Step 2 : Analyze the details given in the question.
The amplitude of the distance between the two points is 2 mm which is half the the amplitude at the center.
We write this as :
½A = A Sin kx
We divide both sides by A to get :
0.5 = Sin 2pie/Lambda × x
Getting Sin inverse of both sides we have :
Sin inverse 0.5 = pie/6
Pie/6 = 2xpie/lambda
Now lambda = 2 L = 2 × 1.5 = 3
Pie/6 = 2xpie/3
Divide both sides by 2 pie/3 to get x
x = pie/6 × 3/2 = 1/4
x = 0.25 m
The other part is = 1.5m - 0.25 m = 1.25 m from the other points.
Step 3 : Calculate the distance between the two points.
1.25m - 0.25 m = 1 m
Answer:
Answer:
The distance between the two points of a string whose length is 1.5 m is 1 meter.
Explanation:
Step 1 : write down the relevant equation.
Y = The amplitude of the wave.
Y = ASinKx × Cosbt
Now set t = 0 then we have :
Y = ASinkx
k = 2pie/lambda
Step 2 : Analyze the details given in the question.
The amplitude of the distance between the two points is 2 mm which is half the the amplitude at the center.
We write this as :
½A = A Sin kx
We divide both sides by A to get :
0.5 = Sin 2pie/Lambda × x
Getting Sin inverse of both sides we have :
Sin inverse 0.5 = pie/6
Pie/6 = 2xpie/lambda
Now lambda = 2 L = 2 × 1.5 = 3
Pie/6 = 2xpie/3
Divide both sides by 2 pie/3 to get x
x = pie/6 × 3/2 = 1/4
x = 0.25 m
The other part is = 1.5m - 0.25 m = 1.25 m from the other points.
Step 3 : Calculate the distance between the two points.
1.25m - 0.25 m = 1 m
Explanation: