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PHXI15:WAVES

354999 Two vibrating tuning forks produce progressive waves $y_{1} = 4 \sin (500 π t)$ and $y_{2} = 2 \sin (506 π t)$ . These tuning forks are held near the ear of a person. The person will hear $α$ beats/s with intensity ratio between maxima and minima equal to $β$ . Find the value of $β - α$ .

1 3

2 6

3 9

4 12

Explanation:

beats $= α = \frac{506 π}{2 π} - \frac{500 π}{2 π} = 3$
$β = \frac{I_{max}}{I_{min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{4 + 2}{4 - 2})}^{2} = 9$
$β - α = 6$

PHXI15:WAVES

355000 A tuning fork produces 6 beats/sec with sonometer wire when its tensions are either $169 N$ or $196 N$ . The frequency of that fork is

1

162 H z

2

190 H z

3

200 H z

4

80 H z

Explanation:

If frequency fork is ' $n$ ' then frequency of sonometer wire is $n + 6$ or $n - 6$
$\Rightarrow \frac{n + 6}{n - 6} = \sqrt{\frac{196}{169}} = \frac{14}{13} \Rightarrow 13 n + 78 = 14 n - 84$
$\Rightarrow n = 162 H z$

PHXI15:WAVES

355001 Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

1 80 and 40

2 100 and 50

3 44 and 22

4 72 and 36

Explanation:

supporting img

Using $n_{Last} = n_{First} + 9 \times 4$
Given that $2 n_{first} = n_{last}$
$\Rightarrow n_{first} = 36 n_{last} = 72$

PHXI15:WAVES

355002 Two tuning forks $P$ and $Q$ , when set vibrating, gives 4 beats/s. If a prong of the fork $P$ is waxed, the beats are reduced to 2 per second. If frequency of $Q$ is $250 H z$ , then frequency of $P$ before filing is

1

246 H z

2

254 H z

3 Any of (1) & (2)

4 None of these

PHXI15:WAVES

354999 Two vibrating tuning forks produce progressive waves $y_{1} = 4 \sin (500 π t)$ and $y_{2} = 2 \sin (506 π t)$ . These tuning forks are held near the ear of a person. The person will hear $α$ beats/s with intensity ratio between maxima and minima equal to $β$ . Find the value of $β - α$ .

1 3

2 6

3 9

4 12

Explanation:

beats $= α = \frac{506 π}{2 π} - \frac{500 π}{2 π} = 3$
$β = \frac{I_{max}}{I_{min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{4 + 2}{4 - 2})}^{2} = 9$
$β - α = 6$

PHXI15:WAVES

355000 A tuning fork produces 6 beats/sec with sonometer wire when its tensions are either $169 N$ or $196 N$ . The frequency of that fork is

1

162 H z

2

190 H z

3

200 H z

4

80 H z

Explanation:

If frequency fork is ' $n$ ' then frequency of sonometer wire is $n + 6$ or $n - 6$
$\Rightarrow \frac{n + 6}{n - 6} = \sqrt{\frac{196}{169}} = \frac{14}{13} \Rightarrow 13 n + 78 = 14 n - 84$
$\Rightarrow n = 162 H z$

PHXI15:WAVES

355001 Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

1 80 and 40

2 100 and 50

3 44 and 22

4 72 and 36

Explanation:

supporting img

Using $n_{Last} = n_{First} + 9 \times 4$
Given that $2 n_{first} = n_{last}$
$\Rightarrow n_{first} = 36 n_{last} = 72$

PHXI15:WAVES

355002 Two tuning forks $P$ and $Q$ , when set vibrating, gives 4 beats/s. If a prong of the fork $P$ is waxed, the beats are reduced to 2 per second. If frequency of $Q$ is $250 H z$ , then frequency of $P$ before filing is

1

246 H z

2

254 H z

3 Any of (1) & (2)

4 None of these

PHXI15:WAVES

354999 Two vibrating tuning forks produce progressive waves $y_{1} = 4 \sin (500 π t)$ and $y_{2} = 2 \sin (506 π t)$ . These tuning forks are held near the ear of a person. The person will hear $α$ beats/s with intensity ratio between maxima and minima equal to $β$ . Find the value of $β - α$ .

1 3

2 6

3 9

4 12

Explanation:

beats $= α = \frac{506 π}{2 π} - \frac{500 π}{2 π} = 3$
$β = \frac{I_{max}}{I_{min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{4 + 2}{4 - 2})}^{2} = 9$
$β - α = 6$

PHXI15:WAVES

355000 A tuning fork produces 6 beats/sec with sonometer wire when its tensions are either $169 N$ or $196 N$ . The frequency of that fork is

1

162 H z

2

190 H z

3

200 H z

4

80 H z

Explanation:

If frequency fork is ' $n$ ' then frequency of sonometer wire is $n + 6$ or $n - 6$
$\Rightarrow \frac{n + 6}{n - 6} = \sqrt{\frac{196}{169}} = \frac{14}{13} \Rightarrow 13 n + 78 = 14 n - 84$
$\Rightarrow n = 162 H z$

PHXI15:WAVES

355001 Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

1 80 and 40

2 100 and 50

3 44 and 22

4 72 and 36

Explanation:

supporting img

Using $n_{Last} = n_{First} + 9 \times 4$
Given that $2 n_{first} = n_{last}$
$\Rightarrow n_{first} = 36 n_{last} = 72$

PHXI15:WAVES

355002 Two tuning forks $P$ and $Q$ , when set vibrating, gives 4 beats/s. If a prong of the fork $P$ is waxed, the beats are reduced to 2 per second. If frequency of $Q$ is $250 H z$ , then frequency of $P$ before filing is

1

246 H z

2

254 H z

3 Any of (1) & (2)

4 None of these

PHXI15:WAVES

354999 Two vibrating tuning forks produce progressive waves $y_{1} = 4 \sin (500 π t)$ and $y_{2} = 2 \sin (506 π t)$ . These tuning forks are held near the ear of a person. The person will hear $α$ beats/s with intensity ratio between maxima and minima equal to $β$ . Find the value of $β - α$ .

1 3

2 6

3 9

4 12

Explanation:

beats $= α = \frac{506 π}{2 π} - \frac{500 π}{2 π} = 3$
$β = \frac{I_{max}}{I_{min}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = {(\frac{4 + 2}{4 - 2})}^{2} = 9$
$β - α = 6$

PHXI15:WAVES

355000 A tuning fork produces 6 beats/sec with sonometer wire when its tensions are either $169 N$ or $196 N$ . The frequency of that fork is

1

162 H z

2

190 H z

3

200 H z

4

80 H z

Explanation:

If frequency fork is ' $n$ ' then frequency of sonometer wire is $n + 6$ or $n - 6$
$\Rightarrow \frac{n + 6}{n - 6} = \sqrt{\frac{196}{169}} = \frac{14}{13} \Rightarrow 13 n + 78 = 14 n - 84$
$\Rightarrow n = 162 H z$

PHXI15:WAVES

355001 Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are

1 80 and 40

2 100 and 50

3 44 and 22

4 72 and 36

Explanation:

supporting img

Using $n_{Last} = n_{First} + 9 \times 4$
Given that $2 n_{first} = n_{last}$
$\Rightarrow n_{first} = 36 n_{last} = 72$

PHXI15:WAVES

355002 Two tuning forks $P$ and $Q$ , when set vibrating, gives 4 beats/s. If a prong of the fork $P$ is waxed, the beats are reduced to 2 per second. If frequency of $Q$ is $250 H z$ , then frequency of $P$ before filing is

1

246 H z

2

254 H z

3 Any of (1) & (2)

4 None of these