6 - Wave in a string Questions Answers
A small ring is threaded on an inextensible frictionless cord of length 2l. The ends of the cord are fixed to a horizontal ceiling. In equilibrium, the ring is at a depth h below the ceiling. Now the ring is pulled aside by a small distance in the vertical plane containing the cord and released. If period of small oscillations of the ring is kπ, find k. Given,
.
while measuring the speed of sound by performing a resonance column expt , a student gets d first resonance conditn at a column length of 18 cm during winter . Repeating d same expt during summer , she measure d column length to be x cm for the second resonance. THEN
1)18>X
2)X>54
3) 54>X>36
4) 36>X>18
ANS (2)
AIEEE 2008
since l α λ α v/n n is frequency
on increasing temperature, v increases so λ increases
for first resonance l = λ/4
and for third resonance L = 3λ'/4
and λ' > λ so x will be more than 54
a cylinderical tube open at both ends has a fundamental frequency f in air the tube is dipped vertically in H2O so that half of it is in H2O the fundamental frequency of the air column is
1)f/2
2)3f/4
3)2f
4) f
aieee 2012
for first case l = λ/2 so l = v/2f
for second case l/2 = λ/4 so l/2 = v/4f
so f in both case are same.
QUES TWO IDENTICAL SOUNDS s1 &s2 reach at a point P in phase. resultant loudness at a point P is ndB higher than loudness of s1 the value of n =?
ANS) 6
Let intensity of one source = I
then intensity at a point where two waves meet in phase = I+I+2√I√Icos0 = 4I
now loudness = 10logI/I0 due to a single source
and loudness at the point where 2 waves meet in phase = 10log4I/I0 = 10logI/I0+ 10log4
since 10log4 = 6 so increment = 6
a uniformrope of lenth 10 m and mass 12 kg hangs verticaly from a rigid support . a block of mass 4 kg is attached to the free end of the rope . atrans verse wave of wavelenth 00.03 m is prodused at the lower end . wavelenth of pulse when it reaches upper end will be?
at top point tension in the string = 16g
and at bottom point T = 4g
and v = √(T/m) here m is mass per unit length which is same throughout the rope.
first calculate v at both the points
frequency will be same throughout
so use the formula v = nλ twice
once at bottom and once at top point
then by diving these 2 formulae we will get the result
try it.
a sonometer wire supports a 4kg load and vibrates in fundamental mode with a tuning fork of frequency 416 Hz. the length of the wire between the bridges is now doubled. in order to maintain fundamental mode the load should be changed to
for sonometer wire
in fundamental mode l = λ/2 = v/2f here l, λ and f are length , wavelength and frequency respectively
now v = √(T/µ) here T and µ are tension and mass per unit length
so finally we will get l = √(T/µ)/2f
make two equations by using this formula and given conditions
in the two equations f and µ will be same