Home
Class 12
PHYSICS
A string is clamped at both the ends and...

A string is clamped at both the ends and it is vibrating in its `4^(th)` harmonic. The equation of the stationary wave is `Y=0.3 sin ( 0.157 x) cos (200 pi t)`. The length of the string is: (All quantities are in SI units.)

Text Solution

Verified by Experts

Given y=0.3`sin (0.157x)cos (200pit)`
So k=0.157 and `omega=200pi=2pif`
`:.f=100` Hz and `v=(omega)/2=(200pi)/0.157=4000m//s`
Now using `f=(nv)/(2l)=(4v)/(2l)=(2v)/l` [here n=4]
`:.l=(2v)/f=(2xx4000)/100=80m`
Promotional Banner

Topper's Solved these Questions

  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 15 : (ELECTRIC CHARGES AND FIELDS)|15 Videos
  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 16 : (ELECTROSTATIC POTENTIAL AND CAPACITANCE)|11 Videos
  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER 13: (OSCILLATIONS)|15 Videos
  • ATOMS

    DISHA PUBLICATION|Exercise EXERCISE-2: CONCEPT APPLICATOR|30 Videos
  • COMMUNICATION SYSTEM

    DISHA PUBLICATION|Exercise EXERCISE-2 : Concept Applicator|30 Videos

Similar Questions

Explore conceptually related problems

A string fixed at both ends, oscillate in 4th harmonic. The displacement of particular wave is given as Y=2Asin(5piX)cos(100pit) . Then find the length of the string?

The equation for the vibration of a string fixed at both ends vibrating in its third harmonic is given by y=2cm sin[(0.6cm^-1)x]cos[(500pis^-1)t] . The length of the string is

The equation of a stationary wave is y = 0.04 sin 200 pi t cos (pi x)/(0.3) with all quantities in SI units. What is the speed of the waves superposed ?

The equation for the vibration of a string fixed at both ends vibrating in its second harmonic is given by y=2sin(0.3cm^(-1))xcos((500pis^(-1))t)cm . The length of the string is :

If a string fixed at both ends, vibrates in its fourth harmonic, the wavelength is 15 cm. What is the length of the string ?

A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by y=Asinkxcos(omegat+(pi)/(3)) . All the symbols have their usual meaning. Mass per unit length of the string is 0.5(gm)/(cm) . Given that A=1 cm and omega=100pi(rad)/(sec) Answer the following 2 questions based on information given (use mu^(2)=10) Q. Starting from t=0 , energy of vibration is completely kinetic at time t t , where t is:

A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by y=Asinkxcos(omegat+(pi)/(3)) . All the symbols have their usual meaning. Mass per unit length of the string is 0.5(gm)/(cm) . Given that A=1 cm and omega=100pi(rad)/(sec) Answer the following 2 questions based on information given (use mu^(2)=10) Q. Starting from t=0 , energy of vibration is completely potential at time t , where t is

A string is rigidly tied at two ends and its equation of vibration is given by y = cos(2 pi t)sin(2 pi x) . Then minimum length of string is

The equation of a stationary wave on a string clamped at both ends and vibrating in its third harmonic is given by y=0.5 sin (0.314"x") cos (600pit) where x and y are in cm and t is in sec. What is the length of the string?