Class 10 - Unit # 10 : General Waves Properties - Solved Numericals

Problem 1

What is the wavelength of a radio wave broadcasted by a radio station with a frequency of 1300 kHz? (Where 1k = 10³, and the speed of the radio wave is 3 × 10⁸ m/s). (Ans: 230.7 m)

Given Data:

• Frequency (f) = 1300 kHz = 1300 × 10³ Hz = 1.3 × 10⁶ Hz
• Speed of radio waves (v) = 3 × 10⁸ m/s
• Wavelength (λ) = ?

Solution:

Using the universal wave equation:

v = f · λ

Rearranging the equation to solve for wavelength (λ):

λ =

v

f

Substituting the values:

λ =

3 × 10⁸

1.3 × 10⁶

λ = 2.307 × 10² m

λ = 230.7 m

Result:

The wavelength of the broadcasted radio waves is 230.7 m.

Problem 2

The waves moving in a pond have a wavelength of 1.6 m, and a frequency of 0.80 Hz. Calculate the speed of these water waves. (Ans: 1.28 m/s)

Given Data:

• Wavelength (λ) = 1.6 m
• Frequency (f) = 0.80 Hz
• Speed of wave (v) = ?

Solution:

Using the wave velocity relation:

v = f · λ

Substituting the given parameters:

v = 0.80 × 1.6

v = 1.28 m/s

Result:

The propagation speed of the water waves in the pond is 1.28 m/s.

Problem 3

If 50 waves pass through a point in a rope in 10 seconds, what are the frequency and the period of the wave? If its wavelength is 8 cm, calculate the wave speed. Explain the type of wave produced. (Ans: f = 5 Hz, T = 0.2 s, v = 0.4 m/s, Transverse Wave)

Given Data:

• Number of waves (N) = 50
• Time taken (t) = 10 s
• Wavelength (λ) = 8 cm = 0.08 m
• Frequency (f) = ?
• Time Period (T) = ?
• Wave speed (v) = ?

Solution:

1. Calculate Frequency (f):

Frequency is the number of wave cycles passing a point per second:

f =

N

t

=

50

10

f = 5 Hz

2. Calculate Time Period (T):

The time period is the reciprocal of the frequency:

T =

1

f

=

1

5

T = 0.2 s

3. Calculate Wave Speed (v):

v = f · λ = 5 × 0.08

v = 0.4 m/s

4. Explain the Type of Wave:

The wave produced in the rope is a transverse wave. This is because the particles of the rope vibrate perpendicularly (up and down) to the direction in which the wave travels (horizontally).

Result:

The frequency is 5 Hz, the period is 0.2 seconds, the speed is 0.4 m/s, and the wave is transverse.

Problem 4

A slinky has produced a longitudinal wave. The wave travels at a speed of 40 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions? (Ans: 0.02 m)

Given Data:

• Wave speed (v) = 40 cm/s = 0.4 m/s
• Frequency (f) = 20 Hz
• Minimum separation between consecutive compressions (λ) = ?

Solution:

In a longitudinal wave, the distance between consecutive compressions is equal to one wavelength (λ):

v = f · λ ⇒ λ =

v

f

Substituting values:

λ =

0.4

20

λ = 0.02 m (or 2 cm)

Result:

The minimum separation between the consecutive compressions is 0.02 m.

Problem 5

Suppose a student is generating waves in a slinky. The student's hand makes one complete forth-and-back oscillation in 0.40 s. The wavelength in the slinky is 0.60 m. For this wave, determine:
(a) Period and frequency
(b) Wave speed (Ans: (a) T = 0.4 s, f = 2.5 Hz (b) v = 1.5 m/s)

Given Data:

• Time for one oscillation (Period, T) = 0.40 s
• Wavelength (λ) = 0.60 m

Solution:

(a) Determine Period and Frequency:

The time period is directly given as the time for one full oscillation:

T = 0.40 s

Frequency (f) is calculated as the inverse of time period:

f =

1

T

=

1

0.40

f = 2.5 Hz

(b) Determine Wave Speed (v):

v = f · λ = 2.5 × 0.60

v = 1.5 m/s

Result:

(a) Period is 0.4 s and frequency is 2.5 Hz.
(b) Wave speed is 1.5 m/s.

Problem 6

If 80 compressions pass through a point in a spring in 20 seconds, calculate the frequency and the period. If two consecutive compressions are 8 cm apart, calculate the wave speed. (Ans: f = 4 Hz, T = 0.25 s, v = 0.32 m/s)

Given Data:

• Number of compressions (N) = 80
• Time taken (t) = 20 s
• Separation between consecutive compressions (λ) = 8 cm = 0.08 m
• Frequency (f) = ?
• Time Period (T) = ?
• Wave speed (v) = ?

Solution:

1. Calculate Frequency (f):

f =

N

t

=

80

20

f = 4 Hz

2. Calculate Time Period (T):

T =

1

f

=

1

4

T = 0.25 s

3. Calculate Wave Speed (v):

v = f · λ = 4 × 0.08

v = 0.32 m/s

Result:

The oscillation frequency is 4 Hz, the wave period is 0.25 seconds, and the propagation speed is 0.32 m/s.

Problem 7

Waves on a swimming pool propagate at 0.90 m/s. If you splash the water at one end of the pool, observe the wave go to the opposite end, reflect, and return in 30.0 s. How far away is the other end of the pool? (Ans: 13.5 m)

Given Data:

• Wave speed (v) = 0.90 m/s
• Round-trip travel time (t) = 30.0 s
• Distance to the other end of the pool (d) = ?

Solution:

First, calculate the total distance covered by the wave during its complete round trip:

Total Distance (S) = v × t

S = 0.90 × 30.0 = 27.0 m

Since the wave travels to the other end and comes back, the length of the pool (d) is exactly half of the total round-trip distance:

d =

S

2

=

27.0

2

d = 13.5 m

Result:

The other end of the swimming pool is 13.5 meters away.

Problem 8

A simple oscillating pendulum has a length of 80.0 cm. Calculate its: (a) Period, and (b) Frequency. (When g = 9.8 m/s²) (Ans: T = 1.79 s, f = 0.56 Hz)

Given Data:

• Length of pendulum (l) = 80.0 cm = 0.80 m
• Acceleration due to gravity (g) = 9.8 m/s²
• Time Period (T) = ?
• Frequency (f) = ?

Solution:

(a) Calculate Time Period (T):

The time period formula of a simple pendulum is:

T = 2π √

l

g

Substituting values:

T = 2(3.1416) √

0.80

9.8

T = 6.283 × √0.08163

T = 6.283 × 0.2857

T = 1.795 s

(b) Calculate Frequency (f):

Frequency is the reciprocal of the period:

f =

1

T

=

1

1.795

f = 0.557 Hz ≈ 0.56 Hz

Result:

The time period of the simple pendulum is 1.79 seconds and its oscillation frequency is 0.56 Hz.