A uniform string fixed at both ends is vibrating in 3rd harmonic and equation y - 4 cm sin- 0-8 cm - 1 - cos- 400 - s - 1 t -The length of the vibrating string is

The correct option is C 11.8 m y=4sin #160; ( 0.8x ) cos #160; ( 400π t ) #160; w=400π =2π f f=200 Hz v= w k = 400π× 100 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Wave Speed expression

A uniform str...

Question

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 (c m)$ $sin [(0.8 {c m}^{- 1}) \times] cos [(400 π s^{- 1}) t]$ The length of the vibrating string is

6.75 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12.45 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11.8 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

18.7 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $11.8 m$
$y = 4 sin (0.8 x) cos (400 π t) w = 400 π = 2 π f f = 200 H z v = \frac{w}{k} = \frac{400 π \times 100}{0.8} m / s 3 \cdot \frac{v}{2 l} = f \Rightarrow 200 = \frac{3 \times 400 π \times 100}{2 \times l \times 0.8} \Rightarrow l = \frac{3 \times 400 π}{4 \times 0.8} = \frac{300 π}{0.8} c m = 11.8 m$
Hence,
option $(C)$ is correct answer.

Suggest Corrections

Similar questions

Q. The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by

y = 2 c m s i n [(0.6 c m^{- 1}) x] c o s [500 π s^{- 1} (t)]

. The length of the string is -

Q. The vibrations of a string fixed at both ends are described by the equation

y = (5.00 m m)

sin [(1.57 c m^{- 1}) x] sin [(314 s^{- 1}) t]

. If the length of a string is

10.0 c m

, locate the nodes. How many loops are formed in the vibration?

Q. A horizontal stretched string, fixed at two ends, is vibrating in its

5^{t h}

harmonic according to the equation,

y (x, t) = (0.10 m) sin [(62.8 m^{- 1}) x] cos [(628 s^{- 1}) t]

. Assuming

π = 3.14

, the correct statement(s) is/are :

The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by

$y = (0.4 c m) s i n [(0.314 c m^{- 1}) x] c o s [(600 π s^{- 1}) t] .$

(a) What is the frequency of vibration ? (b) What are the positions of the nodes? (c) What is the length of the string ? (d) What is the waavelength and the speed of two travelling waves that can interfere to give this vibration ?

Q. A horizontal stretched string fixed at both ends, is vibrating in its fifth harmonic according to the equation,

y (x, t) = (0.01 m) sin [(62.8 m^{- 1}) x] cos [(628 s^{- 1}) t]

. Assume

π = 3.14

, which of the following is/are the correct statements.

Join BYJU'S Learning Program

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation y=4(cm) sin[(0.8cm−1)×]cos[(400πs−1)t]The length of the vibrating string is

The correct option is C 11.8my=4sin(0.8x)cos(400πt)w=400π=2πff=200Hzv=wk=400π×1000.8m/s3⋅v2l=f⇒200=3×400π×1002×l×0.8⇒l=3×400π4×0.8=300π0.8cm=11.8mHence,option (C) is correct answer.

A uniform string fixed at both ends is vibrating in 3rd harmonic and equation $y = 4 (c m)$ $sin [(0.8 {c m}^{- 1}) \times] cos [(400 π s^{- 1}) t]$ The length of the vibrating string is