Numerical Problems 10.1: The time period of a simple pendulum is 2 s. What will be its length on the Earth? What will be its length on the Moon if gm = ge / 6? where ge = 10 msᐨ².
Ans.(1.02 m, 0.17 m)
Given Data:
Gravitational Acceleration on Earth = gₑ = 10 msᐨ²
Gravitational Acceleration on Moon = gₘ = gₑ / 6
so,
gₘ = 10 msᐨ² / 6 = 1.667 msᐨ²
Gravitational Acceleration on Earth = gₑ = 10 msᐨ²
Gravitational Acceleration on Moon = gₘ = gₑ / 6
so,
gₘ = 10 msᐨ² / 6 = 1.667 msᐨ²
Required:
Length of the pendulum on the Moon = lₘ = ?
Length of the pendulum on the Moon = lₘ = ?
Solution:
On the Earth:
Formula for Time Period of Pendulum on the Earth
T = 2 ㄫ `\sqrt frac{l_e}{g_e}`
By taking square both sides
T² = 4 ㄫ² `\frac{l_e}{g_e}`
By Using basic mathematics rules, we will separate lₑ as
lₑ = `\frac{T² g_e}{4 ㄫ² }`
Now by putting values of each quantities
lₑ = `\frac{(2)² ｘ 10 msᐨ²}{4 ｘ (3.1416)² }`
By simplifying we get
lₑ = 1.02 m -------------Ans.1
Thus, the length of the pendulum on the Earth ( lₑ ) for a fixed Time Period of 2S is 1.02 m
On the Moon:
The formula for Time Period of Pendulum on the Moon
T = 2 ㄫ `\sqrt frac{l_m}{g_m}`
By taking square both side
T² = 4 ㄫ² `\frac{l_m}{g_m}`
By Using basic mathematics rules, we will separate lₘ as
lₘ = `\frac{T² g_m}{4 ㄫ² }`
Now by putting values of each quantities
lₘ = `\frac{(2)² ｘ 1.667 msᐨ²}{4 ｘ (3.1416)² }`
By simplifying we get
lₘ = 0.17 m --------------------Ans.2
Thus, the length of the pendulum on the Earth ( lₘ) for a fixed Time Period of 2S is 0.17 m
On the Earth:
Formula for Time Period of Pendulum on the Earth
T = 2 ㄫ `\sqrt frac{l_e}{g_e}`
By taking square both sides
T² = 4 ㄫ² `\frac{l_e}{g_e}`
By Using basic mathematics rules, we will separate lₑ as
lₑ = `\frac{T² g_e}{4 ㄫ² }`
Now by putting values of each quantities
lₑ = `\frac{(2)² ｘ 10 msᐨ²}{4 ｘ (3.1416)² }`
By simplifying we get
lₑ = 1.02 m -------------Ans.1
Thus, the length of the pendulum on the Earth ( lₑ ) for a fixed Time Period of 2S is 1.02 m
On the Moon:
The formula for Time Period of Pendulum on the Moon
T = 2 ㄫ `\sqrt frac{l_m}{g_m}`
By taking square both side
T² = 4 ㄫ² `\frac{l_m}{g_m}`
By Using basic mathematics rules, we will separate lₘ as
lₘ = `\frac{T² g_m}{4 ㄫ² }`
Now by putting values of each quantities
lₘ = `\frac{(2)² ｘ 1.667 msᐨ²}{4 ｘ (3.1416)² }`
By simplifying we get
lₘ = 0.17 m --------------------Ans.2
Thus, the length of the pendulum on the Earth ( lₘ) for a fixed Time Period of 2S is 0.17 m
Numerical Problems 10.2: A pendulum of length 0.99 m is taken to the Moon by an astronaut. The period of the pendulum is 4.9 s. What is the value of g on the surface of the Moon?
Ans. (1.63 msᐨ²)
Given Data:
Length of the pendulum on the Moon = lₘ = 0.99 m
Time Period = T = 4.9 s
Length of the pendulum on the Moon = lₘ = 0.99 m
Time Period = T = 4.9 s
Time Period = T = 4.9 s
Required:
Gravitational Acceleration on Moon = gₘ = ?
Gravitational Acceleration on Moon = gₘ = ?
Solution:
The formula for Time Period of Pendulum on the Moon
T = 2 ㄫ `\sqrt frac{l_m}{g_m}`
By taking square both side
T² = 4 ㄫ² `\frac{l_m}{g_m}`
By Using basic mathematics rules, we will separate gₘ as
gₘ = `\frac{ 4 ㄫ² l_m}{T²}`
Now by putting values of each quantities
gₘ = `\frac{ 4 (3.1416)² ｘ 0.99 m}{(4.9 s)²}`
By solving, we get
gₘ = 1.629 msᐨ² --------------Ans.
Thus, the Gravitational Acceleration of the pendulum on the surface of the Moon is 1.629 msᐨ²
By taking square both side
T² = 4 ㄫ² `\frac{l_m}{g_m}`
By Using basic mathematics rules, we will separate gₘ as
gₘ = `\frac{ 4 ㄫ² l_m}{T²}`
Now by putting values of each quantities
gₘ = `\frac{ 4 (3.1416)² ｘ 0.99 m}{(4.9 s)²}`
By solving, we get
gₘ = 1.629 msᐨ² --------------Ans.
Thus, the Gravitational Acceleration of the pendulum on the surface of the Moon is 1.629 msᐨ²
Numerical Problems 10.3: Find the time periods of a simple pendulum of 1 meter length, placed on Earth and on Moon. The value of g on the surface of Moon is 1/6th of its value on Earth, where ge is 10 msᐨ².
Ans. (2 s, 4.9 s)
Given Data:
Length of the pendulum = l = 1 m
Value of gravitational Acceleration on Earth = gₑ = 10 msᐨ²
Value of gravitational Acceleration on Moon = gₘ = gₑ / 6 = 10 msᐨ² / 6 = 1.667 msᐨ²
Value of gravitational Acceleration on Earth = gₑ = 10 msᐨ²
Value of gravitational Acceleration on Moon = gₘ = gₑ / 6
= 10 msᐨ² / 6 = 1.667 msᐨ²
Required:
Time Period of the pendulum on the Earth = Tₑ = ?
Time Period of the pendulum on the Moon =Tₘ = ?
Solution:
On the Earth:
Formula for Time Period of Pendulum on the Earth
Tₑ = 2 ㄫ `\sqrt frac{l}{g_e}`
On the Earth:
Formula for Time Period of Pendulum on the Earth
Now by putting values of each quantities
Tₑ = 2 ｘ 3.1416 ｘ `\sqrt frac{1 m}{10 msᐨ² }`
By simplifying we get
Tₑ = 2 s -----------------Ans.1
Time Period (Tₑ) of the pendulum on the Earth is 2 seconds
On the Moon:
The formula for Time Period of Pendulum on the Moon
Tₘ = 2 ㄫ `\sqrt frac{l}{g_m}`
Now by putting values of each quantities
Tₘ = 2 ｘ 3.1416 ｘ `\sqrt frac{1 m}{1.667 msᐨ² }`
By simplifying we get
Tₘ = 4.9 s -----------------Ans.2Time Period of the pendulum on the Moon is 4.9 seconds
Time Period of the pendulum on the Moon is 4.9 seconds
Numerical Problems 10.4: A simple pendulum completes one vibration in two seconds. Calculate its length, when g = 10.0 msᐨ² . Ans. (1.02 m)
Given Data:
Time Period = T = 2 s (By Definition: Time Period is the time taken in one complete vibration)
Gravitational Acceleration on Moon = g = 10.0 msᐨ²
Required:
Length of the pendulum on the Moon = l = ?
Solution:
The formula for Time Period of Pendulum
T = 2 ㄫ `\sqrt frac{l}{g}`
By taking square both sides
T² = 4 ㄫ² `\frac{l_e}{g_e}`
By Using basic mathematics rules, we will separate length l as
l = `\frac{T² g}{4 ㄫ² }`
Now by putting values of each given quantities
lₑ = `\frac{(2)² ｘ 10 msᐨ²}{4 ｘ (3.1416)² }`
by solving, we get
l = 1.012 m --------------Ans
Length of the pendulum for a time period of 2s will be 1.012 m
Numerical Problems 10.5: If 100 waves pass through a point of a medium in 20 seconds, what is the frequency and the time period of the wave? If its wavelength is 6 cm, calculate the wave speed. Ans. (5 Hz, 0.2 s, 0.3 msᐨ¹ )
Given Data:
As 100 waves passes through a point in 20 Seconds.
So, Number of waves = 100
Time t = 20 s
Wavelength = 𝝀 = 6 cm = 0.06 m (conversion to SI units)
Required:
Frequency = f = ?
Time Period = T= ?
Wave Speed = v = ?
Solution:
To Find Frequency (f):
By Definition: Frequency (f) is the numbers of waves passes through a point in ONE second
By Unitary Method
In 20 Seconds = 100 waves passes through a point
In 1 Second = `\frac{100}{20}` waves passes through a point
In 1 Second = = 5 waves passes through a point
So, Frequency = f = 5 Hz ------------Ans. 1
-------------------------------------------
To Find Time Period (T):
By Definition: Time Period (T) is the time required to complete one vibration.
1st Method: By Unitary Method
100 waves passes through a point = 20 Seconds
1 waves passes through a point
= `\frac {20}{100}` Seconds = 0.2 Seconds
So, Time Period = T = 0.2 s ------------Ans.2
2nd Method: By Formula
We know that Frequency (f) and Time Period (T) are reciprocal of each other i.e.
T = `\frac {1}{f}` = `\frac {1}{5 Hz}`
T= 0.2 s-------------Ans.2
-------------------------------------------
To Find Wave Speed (v):
We know that Wave Speed (v), Frequency (f) and Wavelength (𝝀) are related as
v = f 𝝀 = 5 Hz x 0.06 m
v = 0.3 msᐨ¹-----------Ans.3
Numerical Problems 10.6: A wooden bar vibrating into the water surface in a ripple tank has a frequency of 12 Hz. The resulting wave has a wavelength of 3 cm. What is the speed of the wave? Ans. (0.36 msᐨ¹ )
Given Data:
Frequency = f = 12 Hz
Wavelength = 𝝀 = 3 cm = 0.03 m (conversion to SI units)
Required:
Speed of the wave = v = ?
Solution:
We know that Wave Speed (v), Frequency (f) and Wavelength (𝝀) are related as
v = f 𝝀
v = 12 Hz x 0.03 m
v = 0.36 msᐨ¹ ----------------Ans.
Numerical Problems 10.7: A transverse wave produced on a spring has a frequency of 190 Hz and travels along the length of the spring of 90 m, in 0.5 s.
(a) What is the period of the wave?
(b) What is the speed of the wave?
(c) What is the wavelength of the wave?
Ans. (0.01 s, 180 m s-1 , 0.95 m)
Given Data:
Frequency = f = 190 Hz
Distance = d = 90 m
Time = t = 0.5 s
Required:
(a) Time Period of the wave = T= ?
(b) Wave Speed = v = ?
(c) Wavelength = 𝝀 = ?
Solution:
(a) We know that Frequency (f) and Time Period (T) are reciprocal of each other i.e.
T = `\frac1{f}` = `\frac1{190 Hz}`
T= 0.005 s = 0.01 s ---------------Ans.1
(b) The formula for wave speed is
Wave Speed = v = `\frac{d}{t}`
v = `\frac{90 m}{0.5 s}`
v = 180 msᐨ¹ -----------Ans.2
(c) We know that Wave Speed (v), Frequency (f) and Wavelength (𝝀) are related as
v = f 𝝀
𝝀 = `\frac{v}{f}`
𝝀 = `\frac{180 msᐨ¹}{190 Hz}`
𝝀 = 0.95 m ------------------Ans.3
Numerical Problems 10.8: Water waves in a shallow dish are 6.0 cm long. At one point, the water moves up and down at a rate of 4.8 oscillations per second.
(a) What is the speed of the water waves?
(b) What is the period of the water waves?
Ans. (0.29 m s-1 , 0.21 s)
Given Data:
Distance (Length of wave) = d = 6.0 cm = 0.06 m
Frequency = f = 4.8 Hz
Distance (Length of wave) = d = 6.0 cm = 0.06 m
Frequency = f = 4.8 Hz
Required:
(a) Wave Speed = v = ?
(b) Time Period of the wave = T= ?
(a) Wave Speed = v = ?
(b) Time Period of the wave = T= ?
Solution:
According to given data, to find wave speed, we first have to find the time period so,
(b) We know that Frequency (f) and Time Period (T) are reciprocal of each other i.e.T = `\frac1{f}` = `\frac1{4.8Hz}`
T= 0.21 s -------------Ans.
Now using T = 0.21 s and d = 0.06 m
(a) The formula for wave speed is
Wave Speed = v = `\frac{d}{t}`
v = `\frac{0.06m}{0.21s}`
v = 0.29 m sᐨ¹ --------------Ans.
(b) We know that Frequency (f) and Time Period (T) are reciprocal of each other i.e.
T = `\frac1{f}` = `\frac1{4.8Hz}`
T= 0.21 s -------------Ans.
Now using T = 0.21 s and d = 0.06 m
(a) The formula for wave speed is
Wave Speed = v = `\frac{d}{t}`
v = `\frac{0.06m}{0.21s}`
v = 0.29 m sᐨ¹ --------------Ans.
Numerical Problems 10.9: At one end of a ripple tank 80 cm across, a 5 Hz vibrator produces waves whose wavelength is 40 mm. Find the time the waves need to cross the tank. Ans. (4 s)
Given Data:
Length = d = 80 cm = 0.8 m (conversion to SI units)
Frequency = f = 5 Hz
Wavelength = 𝝀 = 40 mm = 0.04 m
Required:
Time = t = ?
Solution:
or
t = `\frac{d}{v}` ---------------(1)
Here the value of Speed v is unknown and we can find by using the formula
v = f 𝝀 = 5 Hz ｘ 0.04 m = 0.2 m s⁻¹
Thus putting corresponding values in equation (1) we have
t = `\frac{0.8m}{0.2 m s⁻¹}`
t = 4 s -------------Ans.
Numerical Problems 10.10: What is the Wavelength of the radio waves transmitted by an FM station at 90 MHz? where 1M = 106, and speed of radio wave is 3 x 10⁸ msᐨ¹. Ans. (3.33m)
Given Data:
Frequency = f = 90 MHz = 9 𝘅 10⁷ Hz
Speed = v = 3 𝘅 10⁸ m s⁻¹
Time = t = 0.5 s
Required:
Wavelength = 𝝀 = ?
Solution:
We know that Wave Speed (v), Frequency (f) and Wavelength (𝝀) are related as
v = f 𝝀
𝝀 = `\frac{v}{f}`
𝝀 = `\frac{3 ｘ 10⁸ m s⁻¹}{9 ｘ 10⁷ Hz}`
𝝀 = 0.333 ｘ 10⁸⁻⁷ m
𝝀 = 0.333 ｘ 10 m
or
𝝀 = 3.333 m ---------------Ans
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