Tag: sonometer and laws of transverse vibrations

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Questions Related to sonometer and laws of transverse vibrations

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

When tension of a string is increased by 2.5 N, the initial frequency is altered in the ratio of 3:2. The initial tension in the string is 

  1. 6 N

  2. 5 N

  3. 4 N

  4. 2 N
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let initial tension = T.
Then given that:
$2\nu = \sqrt{\frac{T}{\mu}}$ $2\nu$ is the frequency
Upon increasing the tension by 2.5N, the frequency becomes $3\nu$
$3\nu = \sqrt{\frac{T+2.5}{\mu}}$
Dividing the two equations.
$\frac{3}{2} =\sqrt{ \frac{T+2.5}{T}}$
$\frac{9}{4} = \frac{T+2.5}{T}$
$9T = 4T + 10$
$5T = 10$
$T = 2N$
Option d is correct.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A transverse wave on a string is given by $\displaystyle y=A\sin \left [ \alpha x+\beta t+\frac{\pi }{6} \right ]$ If $\displaystyle \alpha =0.56/cm,\beta =12/sec,A=7.5cm $ then find the displacement and velocity of oscillation at x = 1 cm and t = 1 s is

  1. $\displaystyle 4.6cm,46.5cm: s^{-1}$

  2. $\displaystyle 3.75cm,77.94cm: s^{-1}$

  3. $\displaystyle 1.76cm,7.5cm: s^{-1}$

  4. $\displaystyle 7.5cm,75cm: s^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Displacement y = A sin(alpha*x + beta*t + pi/6). At x=1, t=1: y = 7.5 * sin(0.56 + 12 + 0.52) = 7.5 * sin(13.08). Velocity v = dy/dt = A*beta*cos(alpha*x + beta*t + pi/6). Calculating these values yields approximately 3.75cm and 77.94 cm/s.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire under a tension of 10 kg weight is in unison with tuning fork of frequency 320 Hz. To make the wire vibrate in unison with a tuning fork of frequency 256 Hz, the tension should be altered by 

  1. 3.6 kg decreased

  2. 3.6 kg increased

  3. 6.4 kg decreased

  4. 6.4 kg increased
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

frequency in a sonometer is given as $f= \dfrac{v}{2l}\sqrt{\dfrac{T}{\mu}}$


$\dfrac{f _1}{f _2} = \sqrt{\dfrac{T _1}{T _2}}$

$\dfrac{320}{256} = \sqrt{\dfrac{10\times g}{T _2}}$

$\dfrac{5}{4} = \sqrt{\dfrac{10\times g}{T _2}}$

$T _2= \dfrac{16}{25} \times 10 g \ N$

$T _2 = 6.4 kg$

Tension to be decreased by 3.6 kg. 

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A transverse wave is described by the equation $\displaystyle Y=Y _{0}\sin 2\pi \left ( ft-x/\lambda  \right )$. The maximum particle velocity is equal to four times the wave velocity if

  1. $\displaystyle \lambda =\pi Y _{0}/4 $

  2. $\displaystyle \lambda =\pi Y _{0}/2 $

  3. $\displaystyle \lambda =\pi Y _{0} $

  4. $\displaystyle \lambda =2\pi Y _{0} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$y={ Y } _{ 0 }\sin { 2\pi \left( ft-\frac { x }{ \lambda  }  \right)  } $

Maximum particle velocity
${ V } _{ max }=Aw\ \Rightarrow Aw=4{ V } _{ w }(given)\ Aw=4\left( \frac { w }{ k }  \right) \ A=\frac { 4 }{ k } \ A=\frac { 4 }{ { 2\pi  }/{ \lambda  } } \ \Rightarrow \lambda =\frac { 2\pi A }{ 4 } =\frac { \pi A }{ 2 } \ \left[ \lambda =\frac { \pi { Y } _{ 0 } }{ 2 }  \right] $
Hence option (B) is correct

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates at $120 Hz$. The other end passes over a pulley and supports a $1.50 kg$ mass. The linear mass density of the rope is $0.0550 kg/m$. How wavelength and speed  will change if the mass were increased to $3.00 kg$ ?

  1. <span>both decrease &nbsp;by $\sqrt{5}$ times.</span>

  2. <span>both increase by $\sqrt{2}$ times.</span>

  3. <span>both increase by $\sqrt{5}$ times.</span>

  4. <span>both decrease &nbsp;by $\sqrt{2}$ times.</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since the mass becomes twice, the tension in the wire becomes twice.

Hence the speed becomes $\sqrt{2}$ times since it varies with tension as $\sqrt{T}$.
$v=\lambda\nu$
Thus the wavelength also becomes $\sqrt{2}$ times the initial value.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

The velocity of a transverse wave in a stretched wire is $100ms^{-1}$. If the length of wire is doubled and tension in the string is also doubled, the final velocity of  the transverse wave in the wire is

  1. $100 ms^{-1}$

  2. $141.4 ms^{-1}$

  3. $200 ms^{-1}$

  4. $282.8 ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$v=\sqrt{\dfrac{T}{mass\ per\ unit\ length}}=100\ m/s$
when tension is double $ T$ becomes $2T$ and mass per unit length remains same
so, $v _1 = \sqrt{\dfrac{2T}{mass\ per\ unit\ length}}$
$=\sqrt{2}\times\sqrt{\dfrac T{mass\ per\ unit\ length}}$
$=\sqrt{2}\times 100$
$= 141.4 m/s$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

If the vibrations of a string are to be increased by a factor of two, then tension in the string should be made

  1. Twice

  2. Four times

  3. Eight times

  4. Half
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that, $\displaystyle n=\frac {1}{2} \sqrt {\frac {T}{m}} \Rightarrow n \alpha \sqrt {T}$
If tension is increased four times, the frequency will become twice.

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A sonometer wire, with a suspended mass of $M=1 kg$, is in resonance with a given tuning fork. The apparatus is taken to the moon where the acceleration due to gravity is $\dfrac 16$ that on earth. To obtain resonance on the moon, the value of $M$ should be

  1. $1 kg$

  2. $\sqrt{6}$ kg

  3. $6 kg$

  4. $36 kg$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$f \propto \sqrt{T}=\sqrt{mg}$
$\therefore m _{1}g _{1}=m _{2}g _{2}$
$\Rightarrow (1)g=m\left ( \dfrac{g}{6} \right )$
$\Rightarrow m=6kg$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

In an experiment, the string vibrates in $4$ loops when $50 \ gm-wt$ is placed in pan of weight $15 \ gm$. To make the string vibrate in $6$ loops the weight that has to be removed from the pan is approximately :

  1. $72 \ gm$

  2. $36 \ gm$

  3. $21 \ gm$

  4. $29 \ gm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$frequency\propto \dfrac{1}{no\ of\ loops}$
$f _{1}:f _{2}=6:4=3:2$
$f\alpha \sqrt{T}$
$\therefore \dfrac{3}{2}=\sqrt{\dfrac{65}{x}}$
$\dfrac{9}{4}=\dfrac{65}{x}$
$x=\dfrac{260}{9}=29\ gms$
$\therefore$ weight to be removed is $36 gms.$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A uniform wire of length 20 m and weighing 5 kg hangs vertically. If g = 10 m $s^{-2}$, then the speed of transverse waves in the middle of the wire is:

  1. 10 m$s^{-1}$

  2. 10$\sqrt{2}$ m $s^{-1}$

  3. 4 m $s^{-1}$

  4. 2 m $s^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here $\mu=\dfrac{5}{20}kg/m=\dfrac{1}{4}kg/m$
Tension in the middle of wire,
$T=$ weight of half the wire $=\dfrac{5}{2}\times g=\dfrac{5}{2}\times 10N=25N$
As,$v=\sqrt{\dfrac{T}{\mu}}$
$\Rightarrow v=\sqrt{\dfrac{25}{1/4}}=10m/s$